Questions on Combinatorics Counting Principles Permutations
Combinations Pigeonhole Principle
1. How many ways are there to arrange the letters in the word "MATHEMATICS"?
a) 11!
b) 11!/2!
c) 11!/(2!2!)
d) 11!/(2!2!2!)
Answer: d) 11!/(2!2!2!)
Solution: The word "MATHEMATICS" has 11 letters, but some letters are repeated:
M (2 times), A (2 times), T (2 times), and I (1 time).
We use the formula: 11! / (2! * 2! * 2!) to account for the repetitions.
2. In how many ways can 5 people be seated in a row if two particular people must sit together?
a) 4! * 2!
b) 5! * 2!
c) 4! * 2
d) 5!
Answer: c) 4! * 2
Solution: Consider the two people who must sit together as one unit. Now we have 4 units to
arrange (the pair and 3 individuals). This can be done in 4! ways. The two people within the pair can
be arranged in 2! = 2 ways. So, the total number of arrangements is 4! * 2.
3. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?
a) 120
b) 24
c) 625
d) 5!
Answer: a) 120
Solution: This is a permutation problem. We are selecting 4 digits out of 5 without repetition and
order matters. The formula is P(5,4) = 5!/(5-4)! = 5!/1! = 5 * 4 * 3 * 2 = 120.
4. In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to
each other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to each
other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to each
other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to each
other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to each
other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to each
other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
In how many ways can 3 boys and 2 girls be arranged in a line if the girls must not stand next to each
other?
a) 3! * 2! * 4
b) 5!
c) 3! * 2!
d) 4!
Answer: a) 3! * 2! * 4
Solution: First, arrange the 3 boys (3!). Then, there are 4 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 2 of these 4 spaces for the girls
(C(4,2) = 6). Finally, arrange the girls in these spaces (2!). Total: 3! * C(4,2) * 2! = 6 * 6 * 2 = 72.
5. How many ways are there to select a committee of 3 people from a group of 10?
a) 120
b) 720
c) 210
d) 1000
Answer: a) 120
Solution: This is a combination problem. We use the formula C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) /
(3 * 2 * 1) = 120.
6. In how many ways can 8 people be divided into two groups of 4 each?
a) 35
b) 70
c) 1680
d) 40320
Answer: b) 70
Solution: This is equivalent to choosing 4 people out of 8 to form one group (the other 4 will
automatically form the second group). So, we calculate C(8,4) = 8! / (4! * 4!) = 70.
7. How many ways are there to distribute 5 identical balls into 3 distinct boxes?
a) 21
b) 15
c) 243
d) 125
Answer: a) 21
Solution: This is a stars and bars problem. The formula is C(n+k-1, k-1) where n is the number of
identical objects and k is the number of distinct boxes. Here, C(5+3-1, 3-1) = C(7,2) = 21.
8. In how many ways can the letters of the word "BANANA" be arranged?
a) 6!
b) 60
c) 360
d) 720
Answer: b) 60
Solution: BANANA has 6 letters, but A appears 3 times and N appears 2 times. We use the formula:
6! / (3! * 2!) = 720 / 12 = 60.
9. How many 5-letter words (with or without meaning) can be formed using the letters of the word
"EQUATION" without repetition?
a) 40320
b) 6720
c) 30240
d) 120
Answer: c) 30240
Solution: "EQUATION" has 8 distinct letters. We need to choose 5 out of these 8 letters and then
arrange them. This is P(8,5) = 8! / (8-5)! = 8! / 3! = 40320.
10. In how many ways can 7 people be seated around a circular table?
a) 5040
b) 720
c) 360
d) 5760
Answer: b) 720
Solution: For circular permutations, we use the formula (n-1)! where n is the number of people.
Here, (7-1)! = 6! = 720.
11. How many ways are there to select 3 cards from a standard deck of 52 cards?
a) 22100
b) 140400
c) 15600
d) 52
Answer: a) 22100
Solution: This is a combination problem. We use C(52,3) = 52! / (3! * 49!) = (52 * 51 * 50) / (3 * 2 *
1) = 22100.
12. In how many ways can 4 boys and 3 girls stand in a line so that no two girls are next to each
other?
a) 720
b) 144
c) 2880
d) 5040
Answer: c) 2880
Solution: First, arrange the 4 boys (4!). Then, there are 5 spaces where the girls can be inserted
(before the first boy, between boys, after the last boy). Choose 3 of these 5 spaces for the girls
(C(5,3) = 10). Finally, arrange the girls in these spaces (3!). Total: 4! * C(5,3) * 3! = 24 * 10 * 6 = 1440.
13. How many different necklaces can be made using 6 distinct beads?
a) 720
b) 60
c) 360
d) 120
Answer: b) 60
Solution: This is a circular permutation with reflection. We use the formula (n-1)! / 2, where n is
the number of beads. Here, (6-1)! / 2 = 120 / 2 = 60.
14. In how many ways can 8 people be divided into 3 groups of 3, 3, and 2 people?
a) 1680
b) 3360
c) 5040
d) 6720
Answer: b) 3360
Solution: First, choose 3 people for the first group (C(8,3)), then 3 from the remaining 5 for the
second group (C(5,3)), and the remaining 2 form the last group. Total: C(8,3) * C(5,3) = 56 * 10 = 560.
Multiply by 3! to account for the order of groups: 560 * 6 = 3360.
15. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
a) 1296
b) 360
c) 1080
d) 625
Answer: a) 1296
Solution: For each of the 4 positions, we have 6 choices. Using the multiplication principle: 6 * 6 *
6 * 6 = 1296.
16. In how many ways can 5 identical balls be distributed among 3 distinct boxes if each box must
contain at least one ball?
a) 6
b) 10
c) 15
d) 21
Answer: a) 6
Solution: This is a stars and bars problem with restrictions. We need to distribute 2 balls (5 - 3 = 2)
among 3 boxes, as each box already has one ball. The formula is C(n+k-1, k-1) where n=2 and k=3. So,
C(2+3-1, 3-1) = C(4,2) = 6.
17. How many ways are there to arrange the letters in "MISSISSIPPI"?
a) 11!
b) 34650
c) 24024
d) 39916800
Answer: b) 34650
Solution: "MISSISSIPPI" has 11 letters: 4 S's, 4 I's, 2 P's, and 1 M. We use the formula: 11! / (4! * 4!
* 2! * 1!) = 34650.
18. In a group of 30 people, in how many ways can a president, secretary, and treasurer be chosen if
no person can hold more than one position?
a) 24360
b) 27000
c) 30!
d) 4060
Answer: a) 24360
Solution: This is a permutation problem. We're selecting 3 people out of 30 for distinct positions.
P(30,3) = 30! / (30-3)! = 30 * 29 * 28 = 24360.
19. How many different 5-card hands can be dealt from a standard 52-card deck?
a) 2598960
b) 311875200
c) 5200
d) 52!
Answer: a) 2598960
Solution: This is a combination problem. We use C(52,5) = 52! / (5! * 47!) = 2598960.
20. In how many ways can 10 people be seated in two rows of 5 chairs each?
a) 3628800
b) 14400
c) 113400
d) 120
Answer: a) 3628800
Solution: First, choose 5 people for the first row (C(10,5) = 252). Then arrange these 5 people (5! =
120). The remaining 5 people can be arranged in 5! ways in the second row. Total: C(10,5) * 5! * 5! =
252 * 120 * 120 = 3628800.
21. How many ways are there to select a 5-person committee from 8 men and 7 women if the
committee must have at least 2 men and at least 2 women?
a) 3920
b) 6720
c) 5460
d) 4200
Answer: c) 5460
Solution: We can have either 3 men and 2 women, or 2 men and 3 women.
Case 1: 3 men and 2 women: C(8,3) * C(7,2) = 56 * 21 = 1176
Case 2: 2 men and 3 women: C(8,2) * C(7,3) = 28 * 35 = 980
Total: 1176 + 980 = 2156
22. In how many ways can 7 identical balls be distributed among 4 distinct boxes?
a) 120
b) 210
c) 84
d) 35
Answer: a) 120
Solution: This is a stars and bars problem. We use the formula C(n+k-1, k-1) where n is the number
of identical objects and k is the number of distinct boxes. Here, C(7+4-1, 4-1) = C(10,3) = 120.
23. How many ways are there to arrange 8 people in a line if 3 particular people must not stand next
to each other?
a) 40320
b) 21600
c) 30240
d) 25200
Answer: b) 21600
Solution: First, arrange the 5 people who are not restricted (5!). Then, insert the 3 restricted
people into the 6 spaces between and around the other 5 (P(6,3)). Total: 5! * P(6,3) = 120 * 180 =
21600.
24. In how many ways can 6 identical red balls and 4 identical blue balls be arranged in a line?
a) 210
b) 3628800
c) 10!
d) 120
Answer: a) 210
Solution: This is equivalent to choosing 6 positions out of 10 for the red balls (or 4 positions for the
blue balls). We use C(10,6) = C(10,4) = 10! / (6! * 4!) = 210.
25. How many 6-digit numbers are there in which no digit appears more than twice?
a) 123120
b) 720
c) 54000
d) 46656
Answer: c) 54000
Solution: We can choose the digits in 10C6 = 210 ways. For each choice, we need to arrange 6
digits where at most 2 can be the same. This can be done in 6!/(2!*2!) = 180 ways if two digits are
repeated, or 6! = 720 ways if no digit is repeated. Total: 210 * (180 + 720)/2 = 54000.
26. In how many ways can 4 married couples be seated around a circular table if each person must sit
next to their spouse?
a) 144
b) 40320
c) 24
d) 48
Answer: d) 48
Solution: First, arrange the 4 couples (3! = 6 ways, as it's a circular arrangement). Then, for each
couple, the husband and wife can swap positions (2^4 = 16 ways). Total: 6 * 16 = 96. However, we
need to divide by 2 because rotations of the entire arrangement are considered the same. Final
answer: 96/2 = 48.
27. How many ways are there to select 3 different flavors of ice cream from 10 available flavors and
arrange them in a triple-scoop cone?
a) 720
b) 120
c) 1000
d) 360
Answer: a) 720
Solution: This is a permutation problem. We are selecting 3 flavors out of 10 and arranging them.
P(10,3) = 10! / (10-3)! = 10 * 9 * 8 = 720.
28. In how many ways can 5 books be distributed among 3 students if each student must get at least
one book?
a) 150
b) 243
c) 25
d) 120
Answer: b) 243
Solution: First, give each student one book. Then, we need to distribute the remaining 2 books
among 3 students. This
