Problem Set W:
1. There are 3 doors to a lecture room. In how many ways can a lecturer enter and leave the room?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
2. There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door
and leave from another door?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
3. How many possible combinations can a 3-digit safe code have?
(A) 9C3
(B) 9P3
(C) 39
(D) 93
(E) 103
4. Goodwin has 3 different colored pants and 2 different colored shirts. In how many ways can he
choose a pair of pants and a shirt?
(A) 2
(B) 3
(C) 5
(D) 6
(E) 12
5. In how many ways can 2 doors be selected from 3 doors?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
6. In how many ways can 2 doors be selected from 3 doors for entering and leaving a room?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
7. In how many ways can a room be entered and exited from the 3 doors to the room?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
354 GMAT Math Bible
8. There are 5 doors to a lecture room. Two are red and the others are green. In how many ways can a
lecturer enter the room and leave the room from different colored doors?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
9. Four pool balls—A, B, C, D—are randomly arranged in a straight line. What is the probability that
the order will actually be A, B, C, D ?
(A) 1/4
(B)
1
4C4
(C)
1
4 P4
(D) 1/2!
(E) 1/3!
10. A basketball team has 11 players on its roster. Only 5 players can be on the court at one time. How
many different groups of 5 players can the team put on the floor?
(A) 511
(B) 11C5
(C) 11P5
(D) 115
(E) 11! �� 5!
11. How many different 5-letter words can be formed from the word ORANGE using each letter only
once?
(A) 6P6
(B) 36
(C) 6C6
(D) 66
(E) 6P5
12. How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?
(A) 5!
(B) 5P3
(C) 5C5
2!
(D) 5P5
2!��3!
(E) 5C5
2!��3!
13. How many different six-digit numbers can be formed using all of the following digits:
3, 3, 4, 4, 4, 5
(A) 10
(B) 20
(C) 30
(D) 36
(E) 60
Permutations & Combinations 355
14. This is how Edward’s Lotteries work. First, 9 different numbers are selected. Tickets with exactly 6
of the 9 numbers randomly selected are printed such that no two tickets have the same set of numbers.
Finally, the winning ticket is the one containing the 6 numbers drawn from the 9 randomly. There is
exactly one winning ticket in the lottery system. How many tickets can the lottery system print?
(A) 9P6
(B) 9P3
(C) 9C9
(D) 9C6
(E) 69
15. How many different strings of letters can be made by reordering the letters of the word SUCCESS?
(A) 20
(B) 30
(C) 40
(D) 60
(E) 420
16. A company produces 8 different types of candies, and sells the candies in gift packs. How many
different gift packs containing exactly 3 different candy types can the company put on the market?
(A) 8C2
(B) 8C3
(C) 8P2
(D) 8P3
(E) 8P3
2!
17. Fritz is taking an examination that consists of two parts, A and B, with the following instructions:
Part A contains three questions, and a student must answer two.
Part B contains four questions, and a student must answer two.
Part A must be completed before starting Part B.
In how many ways can the test be completed?
(A) 12
(B) 15
(C) 36
(D) 72
(E) 90
18. A menu offers 2 entrees, 3 main courses, and 3 desserts. How many different combinations of dinner
can be made? (A dinner must contain an entrée, a main course, and a dessert.)
(A) 12
(B) 15
(C) 18
(D) 21
(E) 24
356 GMAT Math Bible
19. In how many ways can 3 red marbles, 2 blue marbles, and 5 yellow marbles be placed in a row?
(A) 3!��2!��5!
(B)
12!
10!
(C)
10!
3!
��
10!
2!
��
10!
5!
(D)
10!
3!��2!��5!
(E)
10!
(3!��2!��5!)2
20. The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the
10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is
the probability that 3 of Sam’s 4 funds grew by at least 10% over last year?
(A) 6C3
10C4
(B) 6C3��4 C1
10C4
(C) 6C3��4 C1
10P4
(D) 6P3��4 P1
10C4
(E) 6P3��4 P1
10P4
21. The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the
10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is
the probability that at least 3 of Sam’s 4 funds grew by at least 10% over the last year?
(A) 6C3
10C4
(B) 6C3��4 C1
10C4
(C) 6C3��4 C1+6C4
10P4
(D) 6P3��4 P1
10C4
(E) 6C3 �� 4C1 + 6C4
10C4
22. In how many ways can the letters of the word ACUMEN be rearranged such that the vowels always
appear together?
(A) 3!��3!
(B)
6!
2!
(C)
4!��3!
2!
(D) 4!��3!
(E)
3!��3!
2!
Permutations & Combinations 357
23. In how many ways can the letters of the word ACCLAIM be rearranged such that the vowels always
appear together?
(A)
7!
2!��2!
(B)
4!��3!
2!��2!
(C)
4!��3!
2!
(D)
5!
2!��2!
(E)
5!
2!
��
3!
2!
24. In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s
appear together?
(A)
8!
3!��2!��2!
(B)
8!
2!��2!
(C)
8!��3!
2!��2!
(D)
8!
2!��3!
(E)
10!
3!��2!��2!
25. In how many ways can the letters of the word GOSSAMERE be rearranged such that all S’s and M’s
appear in the middle?
(A)
9!
2!��2!
(B) 7P6
2!��2!
(C) 7P6
2!
�� 3P3
2!
(D) 6P6
2!
�� 3P3
2!
(E) 10P6
2!
�� 3P3
2!
26. How many different four-letter words can be formed (the words need not be meaningful) using the
letters of the word GREGARIOUS such that each word starts with G and ends with R?
(A) 8P2
(B) 8P2
2!��2!
(C) 8P4
(D) 8P4
2!��2!
(E) 10P2
2!��2!
358 GMAT Math Bible
27. A coin is tossed five times. What is the probability that the fourth toss would turn a head?
(A)
1
5P3
(B)
1
5P9
(C)
1
2
(D)
1
2!
(E)
1
23
28. In how many of ways can 5 balls be placed in 4 tins if any number of balls can be placed in any tin?
(A) 5C4
(B) 5P4
(C) 54
(D) 45
(E) 55
29. On average, a sharpshooter hits the target once every 3 shots. What is the probability that he will hit
the target in 4 shots?
(A) 1
(B) 1/81
(C) 1/3
(D) 65/81
(E) 71/81
30. On average, a sharpshooter hits the target once every 3 shots. What is the probability that he will not
hit the target until 4th shot?
(A) 1
(B) 8/81
(C) 16/81
(D) 65/81
(E) 71/81
31. A new word is to be formed by randomly rearranging the letters of the word ALGEBRA. What is the
probability that the new word has consonants occupying only the positions currently occupied by
consonants in the word ALGEBRA?
(A) 2/120
(B) 1/24
(C) 1/6
(D) 2/105
(E) 1/35
32. Chelsea has 5 roses and 2 jasmines. A bouquet of 3 flowers is to be formed. In how many ways can it
be formed if at least one jasmine must be in the bouquet?
(A) 5
(B) 20
(C) 25
(D) 35
(E) 40
Permutations & Combinations 359
33. In how many ways can 3 boys and 2 girls be selected from a group of 6 boys and 5 girls?
(A) 10
(B) 20
(C) 50
(D) 100
(E) 200
34. In how many ways can a committee of 5 members be formed from 4 women and 6 men such that at
least 1 woman is a member of the committee?
(A) 112
(B) 156
(C) 208
(D) 246
(E) 252
35. In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the
beginning and at the end?
(A)
3!
5!
�� 7!
(B)
5!
6!
�� 7!
(C)
5!
3!
�� 7!
(D)
3!
5!
�� 7!
(E)
5!
7!
�� 7!
36. In how many ways can the letters of the word MAXIMA be arranged such that all vowels are
together?
(A) 12
(B) 18
(C) 30
(D) 36
(E) 72
37. In how many ways can the letters of the word MAXIMA be arranged such that all vowels are together
and all consonants are together?
(A) 12
(B) 18
(C) 30
(D) 36
(E) 42
38. In how many ways can 4 boys and 4 girls be arranged in a row such that no two boys and no two girls
are next to each other?
(A) 1032
(B) 1152
(C) 1254
(D) 1432
(E) 1564
360 GMAT Math Bible
39. In how many ways can 4 boys and 4 girls be arranged in a row such that boys and girls alternate their
positions (that is, boy girl)?
(A) 1032
(B) 1152
(C) 1254
(D) 1432
(E) 1564
40. The University of Maryland, University of Vermont, and Emory University have each 4 soccer
players. If a team of 9 is to be formed with an equal number of players from each university, how
many number of ways can the selections be done?
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
41. In how many ways can 5 persons be seated around a circular table?
(A) 5
(B) 24
(C) 25
(D) 30
(E) 120
42. In how many ways can 5 people from a group of 6 people be seated around a circular table?
(A) 56
(B) 80
(C) 100
(D) 120
(E) 144
43. What is the probability that a word formed by randomly rearranging the letters of the word ALGAE is
the word ALGAE itself?
(A) 1/120
(B) 1/60
(C) 2/7
(D) 2/5
(E) 1/30
=======================================================
Answers and Solutions to Problem Set W
1. There are 3 doors to a lecture room. In how many ways can a lecturer enter and leave the room?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
Recognizing the Problem:
1) Is it a permutation or a combination problem?
Here, order is important. Suppose A, B, and C are the three doors. Entering by door A and leaving by door
B is not the same way as entering by door B and leaving by door A. Hence, AB �� BA implies the problem
is a permutation (order is important).
2) Are repetitions allowed?
Since the lecturer can enter and exit through the same door, repetition is allowed.
3) Are there any indistinguishable objects in the base set?
Doors are different. They are not indistinguishable, so no indistinguishable objects.
Hence, we have a permutation problem, with repetition allowed and no indistinguishable objects.
Method I (Using known formula for the scenario):
Apply Formula 1, nr, from the Formula section. Here, n = 3, (three doors to choose from), r = 2, 2 slots
(one for entry door, one for exit door).
Hence, nr = 32 = 9, and the answer is (D).
Method II (Model 2):
The lecturer can enter the room in 3 ways and exit in 3 ways. So, in total, the lecturer can enter and leave
the room in 9 (= 3 �� 3) ways. The answer is (D). This problem allows repetition: the lecturer can enter by a
door and exit by the same door.
Method III (Model 3):
Let the 3 doors be A, B, and C. We must choose 2 doors: one to enter and one to exit. This can be done in 6
ways: {A, A}, {A, B}, {B, B}, {B, C}, {C, C}, and {C, A}. Now, the order of the elements is important
because entering by A and leaving by B is not same as entering by B and leaving by A. Let’s permute the
combinations, which yields
A – A
A – B and B – A
{B, B}
B – C and C – B
C – C
C – A and A – C
The total is 9, and the answer is (D).
362 GMAT Math Bible
2. There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door
and leave from another door?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
This problem is the same as the previous one, except entering and leaving must be done by different doors
(since the doors are different, repetition is not allowed).
Hence, we have a permutation (there are two slots individually defined: one naming the entering door and
one naming the leaving door), without repetition, and no indistinguishable objects (doors are different).
Recognizing the Problem:
1) Is it a permutation or a combination problem?
Here, order is important. Suppose A, B, and C are the three doors. Entering by door B and leaving by door
C is not same as entering by door C and leaving by door B. Hence, BC �� CB implies the problem is a
permutation (order is important).
2) Are repetitions allowed?
We must count the number of possibilities in which the lecturer enters and exits by different doors, so
repetition is not allowed.
3) Are there any indistinguishable objects in the base set?
Doors are different. They are not indistinguishable, so no indistinguishable objects.
Hence, we have a permutation problem, with repetition not allowed and no indistinguishable objects.
Method I (Using known formula for the scenario):
Apply Formula 2, nPr, from the Formula section. Here, n = 3 (three doors to choose), r = 2 slots (one for
entry door, one for exit door) to place them in.
The calculation is
nPr = 3P2 =
3!
(3�� 2)!
=
3!
1!
=
3�� 2 ��1
1
=
6
The answer is (C).
Method II (Model 2):
The lecturer can enter the room in 3 ways and exit in 2 ways (not counting the door entered). Hence, in
total, the number of ways is 3 �� 2 = 6 (by Model 2) or by the Fundamental Principle of Counting 2 + 2 + 2 =
6. The answer is (C). This is a problem with repetition not allowed.
Method III (Model 3):
Let the 3 doors be A, B, and C. Hence, the base set is {A, B, C}. We have to choose 2 doors—one to enter
and one to exit. This can be done in 3 ways: {A, B}, {B, C}, {C, A} [The combinations {A, A}, {B, B},
and {C, C} were eliminated because repetition is not allowed]. Now, the order of the permutation is
Permutations & Combinations 363
important because entering by A and leaving by B is not considered same as entering by B and leaving by
A. Let’s permute the combinations:
{A, B}
{A, C}
{B, A}
{B, C}
{C, A}
{C, B}
The total is 6, and the answer is (C).
3. How many possible combinations can a 3-digit safe code have?
(A) 9C3
(B) 9P3
(C) 39
(D) 93
(E) 103
The safe combination could be 433 or 334; the combinations are the same, but their ordering is different.
Since order is important for the safe combinations, this is a permutation problem.
A safe code can be made of any of the numbers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. No two objects in the set are
indistinguishable. Hence, the base set does not have any indistinguishable objects.
Repetitions of numbers in the safe code are possible. For example, 334 is a possible safe code.
Hence, the problem is a permutation, with repetition and no indistinguishable objects. Hence, use Formula
1, nr [here, n = 10, r = 3]. The number of codes is 103 = 1000. The answer is (E).
Safe codes allow 0 to be first digit. Here, the same arrangement rules apply to each of the 3 digits. So, this
is a uniform arrangement problem. We can use any formula or model here. But there are non-uniform
arrangement problems. For example, if you are to form a 3-digit number, the first digit has an additional
rule: it cannot be 0 (because in this case the number would actually be 2-digit number). In such scenarios,
we need to use model I or II. The number of ways the digits can be formed by model II is
9 �� 10 �� 10 = 900
4. Goodwin has 3 different colored pants and 2 different colored shirts. In how many ways can he
choose a pair of pants and a shirt?
(A) 2
(B) 3
(C) 5
(D) 6
(E) 12
Model 2:
The pants can be selected in 3 ways and the shirt in 2 ways. Hence, the pair can be selected in 3 �� 2 = 6
ways. The answer is (D).
364 GMAT Math Bible
5. In how many ways can 2 doors be selected from 3 doors?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
It appears that order is not important in this problem: the doors are mentioned but not defined. Also, since
we are selecting doors, it is a combination problem.
The base set is the 3 doors [n = 3]. The doors are different, so there are no indistinguishable objects in the
base set.
The arranged sets are the 2 doors [r = 2] we select. A door cannot be selected twice because “we select 2
doors” clearly means 2 different doors.
Hence, the problem is a combination, with no indistinguishable objects and no repetitions. Hence, using
Formula 3, nCr, yields
3C2 =
n!
r!(n �� r)!
=
3!
2!(3�� 2)!
=
3!
2!��1!
=
3�� 2 ��1
(2 ��1) ��1
=
3
The answer is (B).
6. In how many ways can 2 doors be selected from 3 doors for entering and leaving a room?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
The problem statement almost ended at “3 doors.” The remaining part “entering and leaving” only explains
the reason for the selection. Hence, this does not define the slots. So, this is a combination problem.
Moreover, we are asked to select, not to arrange. Hence, the problem is a combination, with no
indistinguishable objects and no repetitions allowed. Using Formula 3, the number of ways the room can be
entered and left is
nCr=3C2 =
n!
r!(n �� r)!
=
3!
2!(3�� 2)!
=
3!
2!��1!
=
3�� 2 ��1
(2 ��1) ��1
=
3
The answer is (B).
Permutations & Combinations 365
7. In how many ways can a room be entered and exited from the 3 doors to the room?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
There is specific stress on “entered” and “exited” doors. Hence, the problem is not combinational; it is a
permutation (order/positioning is important).
The problem type is “no indistinguishable objects and repetitions allowed”. Hence, by Formula 1, the
number of ways the room can be entered and exited is nr = 32 = 9. The answer is (C).
8. There are 5 doors to a lecture room. Two are red and the others are green. In how many ways can a
lecturer enter the room and leave the room from different colored doors?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
There are 2 red and 3 green doors. We have two cases:
The room can be entered from a red door (2 red doors, so 2 ways) and can be left from a green door (3
green doors, so 3 ways): 2 �� 3 = 6.
The room can be entered from a green door (3 green doors, so 3 ways) and can be left from a red door
(2 red doors, so 2 ways): 3 �� 2 = 6.
Hence, the total number of ways is
2 �� 3 + 3 �� 2 = 6 + 6 = 12
The answer is (E).
9. Four pool balls—A, B, C, D—are randomly arranged in a straight line. What is the probability that
the order will actually be A, B, C, D ?
(A) 1/4
(B)
1
4C4
(C)
1
4 P4
(D) 1/2!
(E) 1/3!
This is a permutation problem (order is important).
A ball cannot exist in two slots, so repetition is not allowed.
Each ball is given a different identity A, B, C, and D, so there are no indistinguishable objects.
Here, n = 4 (number of balls to arrange) in r = 4 (positions). We know the problem type, and the formula to
use. Hence, by Formula 2, the number of arrangements possible is 4P4, and {A, B, C, D} is just one of the
arrangements. Hence, the probability is 1 in 4P4, or
1
4 P4
. The answer is (C).
366 GMAT Math Bible
10. A basketball team has 11 players on its roster. Only 5 players can be on the court at one time. How
many different groups of 5 players can the team put on the floor?
(A) 511
(B) 11C5
(C) 11P5
(D) 115
(E) 11! �� 5!
The task is only to select a group of 5, not to order them. Hence, this is a combination problem. There are
11 players; repetition is not possible among them (one player cannot be counted more than once); and they
are not given the same identity. Hence, there are no indistinguishable objects. Using Formula 3, groups of 5
can be chosen from 11 players in 11C5 ways. The answer is (B).
11. How many different 5-letter words can be formed from the word ORANGE using each letter only
once?
(A) 6P3
(B) 36
(C) 6C6
(D) 66
(E) 6P5
In the problem, order is important because ORGAN is a word formed from ORANGE and ORNAG is a
word formed from ORANGE, but they are not the same word. Repetition is not allowed, since each letter in
the original word is used only once.
The problem does not have indistinguishable objects because no two letters of the word ORANGE are the
same. Hence, by Formula 2, nPr, the answer is 6P5, which is choice (E).
12. How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?
(A) 5!
(B) 5P3
(C) 5C5
2!
(D) 5P5
2!��3!
(E) 5C5
2!��3!
The word “unequal” indicates that this is a permutation problem, because 11532 is the same combination as
11235, but they are not equal. Hence, they are permutations, different arrangements in a combination.
The indistinguishable objects in the base set {1, 1, 2, 3, 5} are the two 1’s.
Since each digit of the number 11235 (objects in the base set) is used only once, repetitions are not allowed.
Hence, by Formula 4, the number of unequal 5-digit numbers that can be formed is
5P5
2!
=
5!
0!��5!
2!
=
5P0
2!
The answer is (B).
Permutations & Combinations 367
13. How many different six-digit numbers can be formed using all of the following digits:
3, 3, 4, 4, 4, 5
(A) 10
(B) 20
(C) 30
(D) 36
(E) 60
Forming a six-digit number is a permutation because the value of the number changes with the different
arrangements.
Since we have indistinguishable numbers in the base set, the regular permutations generate repeating
numbers. But we are asked for only different six-digit numbers. So, we count only 1 for each similar
permutation.
There are two sets of indistinguishable objects in the base set: two 3’s and three 4’s.
No repetitions are allowed since all elements in the base set are to be used in each number.
Hence, by Formula 4, the formula for permutations with no repetitions and with distinguishable objects, the
number of six-digit numbers that can be formed is
6P6
2!��3!
=
6!
2!��3!
= 60
The answer is (E).
14. This is how Edward’s Lotteries work. First, 9 different numbers are selected. Tickets with exactly 6
of the 9 numbers randomly selected are printed such that no two tickets have the same set of numbers.
Finally, the winning ticket is the one containing the 6 numbers drawn from the 9 randomly. There is
exactly one winning ticket in the lottery system. How many tickets can the lottery system print?
(A) 9P6
(B) 9P3
(C) 9C9
(D) 9C6
(E) 69
The only condition is that the winning ticket has the same set of numbers as the drawn numbers (in
whatever order). Hence, order is not important. Now, count the combinations.
Since the numbers in the base set (9 numbers) are different, the base set does not have indistinguishable
objects.
Six of the 9 different numbers are selected by the lottery system, so no repetitions are allowed.
By Formula 3, the formula for combinations with no repetitions and no indistinguishable objects, the
number of possible selections by the lottery system is 9C6.
Since only one winning ticket (winning combination) exists per lottery system, there is of 9C6 tickets per
lottery system. The answer is (D).
368 GMAT Math Bible
15. How many different strings of letters can be made by reordering the letters of the word SUCCESS?
(A) 20
(B) 30
(C) 40
(D) 60
(E) 420
The word SUCCESS is a different word from SUSSECC, while they are the same combination. Hence, this
is a permutation problem, not a combination problem.
There are two sets of indistinguishable objects in the base set: 2 C’s and 3 S’s.
Each letter is used only once in each reordering (so do not allow repetition).
Hence, we have a permutation problem, with indistinguishable objects and no repetitions. Using Formula 4,
the formula for permutations with no repetitions but with distinguishable objects in the Formula section,
yields n = 7 (base word has 7 letters), and r = 7 (each new word will have 7 letters). The repetitions are 2
C’s and 3 S’s. Hence, the total number of permutations is
7P7
2!��3!
=
7 �� 6 �� 5�� 4 �� 3�� 2 ��1
2 �� 6
=
7 �� 5�� 4 �� 3 =
420
The answer is (E).
16. A company produces 8 different types of candies, and sells the candies in gift packs. How many
different gift packs containing exactly 3 different candy types can the company put on the market?
(A) 8C2
(B) 8C3
(C) 8P2
(D) 8P3
(E) 8P3
2!
The phrase “8 different candies” indicates the base set does not have indistinguishable objects.
Since no placement slots are defined, this is a combination problem. We need only to choose 3 of 8
candies; we do not need to order them.
Repetitions are not allowed in the sets formed.
By Formula 3, the formula for permutations with no repetitions but with distinguishable objects yields 8C3,
which is Choice (B).
Permutations & Combinations 369
17. Fritz is taking an examination that consists of two parts, A and B, with the following instructions:
Part A contains three questions, and a student must answer two.
Part B contains four questions, and a student must answer two.
Part A must be completed before starting Part B.
In how many ways can the test be completed?
(A) 12
(B) 15
(C) 36
(D) 72
(E) 90
The problem has two parts.
Each part is a permutation problem with no indistinguishable objects (no 2 questions are the same in either
part), and repetitions are not allowed (the same question is not answered twice).
Hence, the number of ways of answering the first part is 3P2 (2 questions to answer from 3), and the number
of ways of answering the second part is 4P2 (2 questions to answer from 4).
By the Fundamental Principle of Counting, the two parts can be done in
3P2 �� 4P2 = 6 �� 12 = 72 ways
The answer is (D).
Method II [Model 2]:
The first question in Part A can be chosen to be one of the 3 questions in Part A.
The second question in Part A can be chosen to be one of the remaining 2 questions in Part A.
The first question in Part B can be chosen to be one of the 4 questions in Part B.
The second question in Part B can be chosen to be one of the remaining 3 questions in Part B.
Hence, the number of choices is
3 �� 2 �� 4 �� 3 = 72
The answer is (D).
Method III [Fundamental Principle of Counting combined with Model 2]:
The first question in part A can be chosen to be one of the 3 questions in Part A.
The second question in part A can be chosen to be one of the 3 questions in Part A allowing the repetitions.
Hence, number of permutations is 3 �� 3 = 9. There are 3 repetitions [Q1 & Q1, Q2 & Q2, Q3 & Q3]. The
main question does not allow repetitions since you would not answer the same question again. Deleting
them, we have 9 – 3 = 6 ways for Part A.
The first question in Part B can be chosen to be one of the 4 questions in Part B.
The second question in Part B can be chosen to be one of 4 questions in Part B. Hence, the number of
permutations is 4 �� 4 = 16. There are 4 repetitions [Q1 & Q1, Q2 & Q2, Q3 & Q3, Q4 & Q4]. The main
question does not allow repetitions since you would not answer the same question again. Deleting them, we
have 16 – 4 = 12 ways for Part A.
Hence, the number of choices is
6 �� 12 = 72
The answer is (D).
370 GMAT Math Bible
18. A menu offers 2 entrees, 3 main courses, and 3 desserts. How many different combinations of dinner
can be made? (A dinner must contain an entrée, a main course, and a dessert.)
(A) 12
(B) 15
(C) 18
(D) 21
(E) 24
The problem is a mix of 3 combinational problems. The goal is to choose 1 of 2 entrees, then 1 of 3 main
courses, then 1 of 3 desserts. The choices can be made in 2, 3, and 3 ways, respectively. Hence, the total
number of ways of selecting the combinations is 2 �� 3 �� 3 = 18. The answer is (C).
We can also count the combinations by the Fundamental Principle of Counting:
The Fundamental Principle of Counting states:
The total number of possible outcomes of a series of decisions, making selections from various categories,
is found by multiplying the number of choices for each decision.
Counting the number of choices in the final column above yields 18.
Dessert 1
Dessert 2
Entrée 1
Entrée 2
Dessert 3
Main Course 1
Main Course 2
Main Course 3
Main Course 1
Main Course 2
Main Course 3
Dessert 1
Dessert 2
Dessert 3
Dessert 1
Dessert 2
Dessert 3
Dessert 1
Dessert 2
Dessert 3
Dessert 1
Dessert 2
Dessert 3
Dessert 1
Dessert 2
Dessert 3
Total 18
Permutations & Combinations 371
19. In how many ways can 3 red marbles, 2 blue marbles, and 5 yellow marbles be placed in a row?
(A) 3!��2!��5!
(B)
12!
10!
(C)
10!
3!
��
10!
2!
��
10!
5!
(D)
10!
3!��2!��5!
(E)
10!
(3!��2!��5!)2
Since the question is asking for the number of ways the marbles can be placed adjacent to each other, this is
a permutation problem.
The base set has 3 red marbles (indistinguishable objects), 2 blue marbles (indistinguishable objects) and 5
yellow marbles (indistinguishable objects). The possible arrangements are 3 + 2 + 5 = 10 positions.
The same marble cannot be used twice, so no repetitions are allowed. Formula 4, nPr, and the method for
indistinguishable objects (that is, divide the number of permutations, nPr, by the factorial count of each
indistinguishable object [see Formulas section]) yield the number of permutations:
10P10
3!��2!��5!
=
10!
3!��2!��5!
The answer is (D).
20. The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the
10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is
the probability that 3 of Sam’s 4 funds grew by at least 10% over last year?
(A) 6C3
10C4
(B) 6C3��4 C1
10C4
(C) 6C3��4 C1
10P4
(D) 6P3��4 P1
10C4
(E) 6P3��4 P1
10P4
There are 6 winning funds that grew more than 10%, and 4 losing funds that grew less than 10%.
The problem can be split into 3 sub-problems:
We have the specific case where Sam must choose 4 funds, 3 of which are winning, so the remaining fund
must be losing. Let’s evaluate the number of ways this can be done. [Note: The order in which the funds
are chosen is not important because whether the first 3 funds are winning and the 4th one is losing, or the
first fund is losing and the last 3 are winning; only 3 of 4 funds will be winning ones. Hence, this is a
combination problem.] The problem has 2 sub-problems:
1. Sam must choose 3 of the 6 winning funds. This can be done in 6C3 ways.
2. Sam must choose one losing fund (say the 4th fund). There are 10 – 6 = 4 losing funds. Hence,
the 4th fund can be any one of the 4 losing funds. The selection can be done in 4C1 ways.
372 GMAT Math Bible
Hence, the total number of ways of choosing 3 winning funds and 1 losing one is 6C3 �� 4C1.
3. Sam could have chosen 4 funds in 10C4 ways.
Hence, the probability that 3 of Sam’s 4 funds grew by at least 10% over last year is
6C3 �� 4C1
10C4
=
20 �� 4
210
=
8
21
The answer is (B).
21. The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the
10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is
the probability that at least 3 of Sam’s 4 funds grew by at least 10% over the last year?
(A) 6C3
10C4
(B) 6C3��4 C1
10C4
(C) 6C3��4 C1+6C4
10P4
(D) 6P3��4 P1
10C4
(E) 6C3 �� 4C1 + 6C4
10C4
There are 6 winning funds that grew more than 10%, and 4 losing funds that grew less than 10%.
The problem can be split into 3 sub-problems:
1) Sam has to choose 3 winning funds. This can be done in 6C3 ways.
2) Sam has to choose 1 losing fund. This can be done in 4C1 ways.
Or
3) Sam has to choose all 4 funds to be winning funds. This can be done in 6C4 ways.
This is how Sam chooses at least 3 winning funds.
Hence, the total number of ways of choosing at least 3 winning funds is 6C3 �� 4C1 + 6C4.
If there were no restrictions (such as choosing at least 3 winning funds), Sam would have chosen funds in
10C4 ways.
Hence, the probability that at least 3 of Sam’s 4 funds grew by at least 10% over the last year is
6C3 �� 4C1 + 6C4
10C4
The answer is (E).
Permutations & Combinations 373
22. In how many ways can the letters of the word ACUMEN be rearranged such that the vowels always
appear together?
(A) 3!��3!
(B)
6!
2!
(C)
4!��3!
2!
(D) 4!��3!
(E)
3!��3!
2!
The word “rearranged” indicates that this is a permutation problem.
The base set {A, C, U, M, E, N} has no indistinguishable objects.
Repetition is not allowed.
Since the 3 vowels must appear together, treat the three as an inseparable unit. Hence, reduce the base set to
{{A, U, E}, C, M, N}. Now, there are 4 different units in the base set, and they can be arranged in 4P4 = 4!
ways. The unit {A, U, E} can itself be internally arranged in 3P3 = 3! ways. Hence, by The Fundamental
Principle of Counting, the total number of ways of arranging the word is 4!��3!. The answer is (D).
23. In how many ways can the letters of the word ACCLAIM be rearranged such that the vowels always
appear together?
(A)
7!
2!��2!
(B)
4!��3!
2!��2!
(C)
4!��3!
2!
(D)
5!
2!��2!
(E)
5!
2!
��
3!
2!
The word “rearranged” indicates that this is a permutation problem.
Since the 3 vowels A, A, and I must appear together, treat the three as an inseparable unit. Hence, reduce
the base set to {{A, A, I}, C, C, L, M}.
The set has two indistinguishable objects, C’s.
Also, repetitions are not allowed since we rearrange the word.
Hence, the number of permutations that can be created with units of the set is 5P5
2!
=
5!
2!
.
Now, let’s see how many permutations we can create with the unit {A, A, I}.
The unit {A, A, I} has two indistinguishable objects, A’s.
Also, repetitions are not allowed.
Hence, by Formula 4, the number of ways of permuting it is 3P3
2!
=
3!
2!
.
374 GMAT Math Bible
Hence, by The Fundamental Principle of Counting, the total number of ways of rearranging the letters is
5!
2!
��
3!
2!
The answer is (E).
24. In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s
appear together?
(A)
8!
3!��2!��2!
(B)
8!
2!��2!
(C)
8!��3!
2!��2!
(D)
8!
2!��3!
(E)
10!
3!��2!��2!
The word “rearranged” indicates that this is a permutation problem.
Since all 3 G’s are together, treat them a single inseparable unit. Hence, the base set reduces to {{G, G, G},
A, R, A, N, T, U, N}. There are 8 independent units, 2 A’s (indistinguishable), and two N’s
(indistinguishable). No unit is used twice, so there are no repetitions. Hence, by Formula 4, the number of
arrangements is 8P8
2!��2!
=
8!
2!��2!
.
The 3 G’s can be rearranged amongst themselves in 3P3
3!
=
3!
3!
= 1 way. Hence, the total number of ways the
letters can be rearranged is
8!
2!��2!
��1 =
8!
2!��2!
The answer is (B).
25. In how many ways can the letters of the word GOSSAMERE be rearranged such that all S’s and M’s
appear in the middle?
(A)
9!
2!��2!
(B) 7P6
2!��2!
(C) 7P6
2!
�� 3P3
2!
(D) 6P6
2!
�� 3P3
2!
(E) 10P6
2!
�� 3P3
2!
The word “rearranged” indicates that this is a permutation problem.
Since S and M must appear in the middle, treat them as an inseparable unit and reserve the middle seat for
them. Correspondingly, bracket them in the base set. The new base set becomes {{S, S, M}, G, O, A, E, R,
E}. Hence, we have the following arrangement:
Permutations & Combinations 375
____ ____ _____ {S, S, M} _____ _____ ______
Now, the remaining 6 units G, O, A, E, R, and E can be arranged in the 6 blank slots; and for each
arrangement, every permutation inside the unit {S, S, M} is allowed.
Hence, the blank slots can be filled in 6P6
2!
(E repeats twice) ways.
And the unit {S, S, M} can be internally arranged in 3P3
2!
ways.
Hence, by Model 2, the total number of ways the letters can be rearranged is
6P6
2!
�� 3P3
2!
The answer is (D).
26. How many different four-letter words can be formed (the words need not be meaningful) using the
letters of the word GREGARIOUS such that each word starts with G and ends with R?
(A) 8P2
(B) 8P2
2!��2!
(C) 8P4
(D) 8P4
2!��2!
(E) 10P2
2!��2!
Place one G in the first slot and one R in the last slot:
G __ __ R
The remaining letters, {G, R, E, A, I, O, U, S}, can be arranged in the remaining 2 slots in 8P2 (no
indistinguishable objects nor repetition). The answer is (A).
Note: Since the two G’s in the base word are indistinguishable, the word G1G2AR is the same as G2G1AR.
Hence, the internal arrangement of the G’s or, for the same reason, the R’s is not important.
27. A coin is tossed five times. What is the probability that the fourth toss would turn a head?
(A)
1
5P3
(B)
1
5P9
(C)
1
2
(D)
1
2!
(E)
1
23
The fourth toss is independent of any other toss. The probability of a toss turning heads is 1 in 2, or simply
1/2. Hence, the probability of the fourth toss being a head is 1/2. The answer is (C).
376 GMAT Math Bible
Method II:
Each toss has 2 outcomes. Hence, 5 tosses have 2 �� 2 �� 2 �� … 2 (5 times) = 25 outcomes (permutation with
repetition over r = 2 and n = 5 [repetitions allowed: the second and the fourth toss may both yield heads or
tails]).
Reserve the third toss for a head. Now, the number of ways the remaining 4 tosses can be tossed is 24
(repetitions allowed). The probability is
24
25 =
1
2
. The answer is (C).
28. In how many of ways can 5 balls be placed in 4 tins if any number of balls can be placed in any tin?
(A) 5C4
(B) 5P4
(C) 54
(D) 45
(E) 55
The first ball can be placed in any one of the four tins.
Similarly, the second, the third, the fourth, and the fifth balls can be placed in any one of the 4 tins.
Hence, the number of ways of placing the balls is 4 �� 4 �� 4 �� 4 �� 4 = 45. The answer is (D).
Note: We used Model 2 here.
29. On average, a sharpshooter hits the target once every 3 shots. What is the probability that he will hit
the target in 4 shots?
(A) 1
(B) 1/81
(C) 1/3
(D) 65/81
(E) 71/81
The sharpshooter hits the target once in 3 shots. Hence, the probability of hitting the target is 1/3. The
probability of not hitting the target is 1 – 1/3 = 2/3.
Now, (the probability of not hitting the target even once in 4 shots) + (the probability of hitting at least once
in 4 shots) equals 1, because these are the only possible cases.
Hence, the probability of hitting the target at least once in 4 shots is
1 – (the probability of not hitting even once in 4 shots)
The probability of not hitting in the 4 chances is
2
3
��
2
3
��
2
3
��
2
3
=
16
81
. Now, 1 – 16/81 = 65/81. The answer is
(D).
This methodology is similar to Model 2. You might try analyzing why. Clue: The numerators of
2
3
��
2
3
��
2
3
��
2
3
=
16
81
are the number of ways of doing the specific jobs, and the denominators are the number
of ways of doing all possible jobs.
Permutations & Combinations 377
30. On average, a sharpshooter hits the target once in every 3 shots. What is the probability that he will
not hit the target until the 4th shot?
(A) 1
(B) 8/81
(C) 16/81
(D) 65/81
(E) 71/81
The sharpshooter hits the target once in every 3 shots. Hence, the probability of hitting the target is 1/3. The
probability of not hitting the target is 1 – 1/3 = 2/3.
He will not hit the target on the first, second, and third shots, but he will hit it on the fourth shot. The
probability of this is
2
3
��
2
3
��
2
3
��
1
3
=
8
81
The answer is (B).
This methodology is similar to Model 2. You might try analyzing why. Clue: The numerators of
2
3
��
2
3
��
2
3
��
2
3
=
16
81
are the number of ways of doing the specific jobs, and the denominators are the number
of ways of doing all possible jobs.
31. A new word is to be formed by randomly rearranging the letters of the word ALGEBRA. What is the
probability that the new word has consonants occupying only the positions currently occupied by
consonants in the word ALGEBRA?
(A) 2/120
(B) 1/24
(C) 1/6
(D) 2/105
(E) 1/35
If we do not put restrictions on the arrangements of the consonants, then by Formula 4 the number of words
that can be formed from the word ALGEBRA is
7!
2!
(A repeats).
If we constrain that the positions of consonants is reserved only for consonants, then the format of the new
arrangement should look like this
A, L, G, E, B, R, A
V, C, C, V, C, C, V
V for vowels, C for consonants.
The 4 slots for consonants can be filled in 4P4 = 4! ways, and the 3 slots for vowels can be filled in
3!
2!
(A
repeats) ways. Hence, by Formula 2, the total number of arrangements in the format is 4!
3!
2!
��
�� ��
��
�� ��
.
Hence, the probability is
4!
3!
2!
��
�� ��
��
�� ��
7!
2!
=
1
35
The answer is (E).
378 GMAT Math Bible
32. Chelsea has 5 roses and 2 jasmines. A bouquet of 3 flowers is to be formed. In how many ways can it
be formed if at least one jasmine must be in the bouquet?
(A) 5
(B) 20
(C) 25
(D) 35
(E) 40
This is a selection problem because whether you choose a jasmine first or a rose first does not matter.
The 3 flowers in the bouquet can be either 1 jasmine and 2 roses or 2 jasmines and 1 rose.
1 of 2 jasmines can be selected in 2C1 ways.
2 of 5 roses can be selected in 5C2 ways.
The subtotal is 2C1 �� 5C2 =
2!
1!��1!
��
5!
3!��2!
= 2 ��10 = 20 .
2 of 2 jasmines can be selected in 2C2 ways.
1 of 5 roses can be selected in 5C1 ways.
The subtotal is 2C2 �� 5C1 =
2!
2!��0!
��
5!
4!��1!
= 1�� 5 = 5.
The grand total is 20 + 5 = 25 ways. The answer is (C).
33. In how many ways can 3 boys and 2 girls be selected from a group of 6 boys and 5 girls?
(A) 10
(B) 20
(C) 50
(D) 100
(E) 200
We have two independent actions to do:
1) Select 3 boys from 6 boys.
2) Select 2 girls from 5 girls.
Selection is a combination problem since selection does not include ordering. Hence, by Model 2, the
number of ways is
(6C3 ways for boys) �� (5C2 ways for girls) =
6!
3!��3!
��
�� ��
��
�� ��
��
5!
2!��3!
��
�� ��
��
�� ��
=
20 ��10 =
200
The answer is (E).
Permutations & Combinations 379
34. In how many ways can a committee of 5 members be formed from 4 women and 6 men such that at
least 1 woman is a member of the committee?
(A) 112
(B) 156
(C) 208
(D) 246
(E) 252
Forming members of committee is a selection action and therefore this is a combination problem. Whether
you select A first and B next or vice versa, it will only be said that A and B are members of the committee.
The number of ways of forming the committee of 5 from 4 + 6 = 10 people is 10C5. The number of ways of
forming a committee with no women (5 members to choose from 6 men) is 6C5. Hence, the number of ways
of forming the combinations is
10C5��6C5 =
10!
5!��5!
��
6!
5!
=
252 �� 6 =
246
The answer is (D).
35. In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the
beginning and at the end?
(A)
3!
5!
�� 7!
(B)
5!
6!
�� 7!
(C)
5!
3!
�� 7!
(D)
3!
5!
�� 7!
(E)
5!
7!
�� 7!
The arrangement is a permutation, and there are no indistinguishable objects because no two boys or girls
are identical. The first and the last slots hold two of the 5 boys, and the remaining slots are occupied by the
4 girls and the 3 remaining boys.
The first and the last slots can hold 2 of the 5 boys in 5P2 ways, and the 3 boys and the 4 girls position
themselves in the middle slots in 7P7 ways. Hence, there are
5!
3!
�� 7! possible arrangements. The answer is
(C).
380 GMAT Math Bible
36. In how many ways can the letters of the word MAXIMA be arranged such that all vowels are
together?
(A) 12
(B) 18
(C) 30
(D) 36
(E) 72
The base set can be formed as {{A, I, A}, M, X, M}. The unit {A, I, A} arranges itself in 3P3
2!
ways. The 4
units in the base set can be arranged in 4P4/2! ways. Hence, the total number of ways of arranging the letters
is
3P3
2!
�� 4 P4
2!
=
3!
2!
��
4!
2!
= 3��12 = 36
The answer is (D).
37. In how many ways can the letters of the word MAXIMA be arranged such that all vowels are together
and all consonants are together?
(A) 12
(B) 18
(C) 30
(D) 36
(E) 42
Since vowels are together and consonants are together, arrange the base set as {{A, I, A}, {M, X, M}}.
Here, {A, I, A} and {M, X, M} are two inseparable units.
The two units can be mutually arranged in 2P2 ways.
Each unit has 3 objects, 2 of which are indistinguishable.
Hence, the number of permutations of each is 3P2
2!
.
Hence, the total number of arrangements possible is
( 2P2) 3P2
2!
��
�� ��
��
�� ��
3P2
2!
��
�� ��
��
�� ��
=
(2)(3)(3) =
18
The answer is (B).
Permutations & Combinations 381
38. In how many ways can 4 boys and 4 girls be arranged in a row such that no two boys and no two girls
are next to each other?
(A) 1032
(B) 1152
(C) 1254
(D) 1432
(E) 1564
Form the base set as {{B1, B2, B3, B4}, {G1, G2, G3, G4}}; Looking at the problem, either B’s or G’s
occupy the odd slots and the other one occupies the even slots. Choosing one to occupy an odd slot set can
be done in 2P1 ways, and the other one is automatically filled by the other group.
Now, fill B’s in odd slots (the number of ways is 4P4), and fill G’s in even slots (the number of ways is 4P4
ways). The total number of ways of doing this is 4P4 �� 4P4.
The number of ways of doing all of this is 2! �� 4P4 �� 4P4 = 2 �� 24 �� 24 = 1152. The answer is (B).
39. In how many ways can 4 boys and 4 girls be arranged in a row such that boys and girls alternate their
positions (that is, boy girl)?
(A) 576
(B) 1152
(C) 1254
(D) 1432
(E) 1564
Form the base set as {{B1, B2, B3, B4}, {G1, G2, G3, G4}}; the set {B1, B2, B3, B4} occupies alternate
positions, as does the set {G1, G2, G3, G4}.
Now there are odd slots and even slots. Each odd slot alternates, and each even slot alternates. Therefore,
we have two major slots: even and odd and two units to occupy them: {B1, B2, B3, B4} and {G1, G2, G3,
G4}. This can be done in 2P1 ways.
An alternate explanation for this is: The person starting the row can be chosen in 2 ways (i.e., either boys
start the first position and arrange alternately or girls start and do the same), either B starts first or G starts
first.
Either way, the positions {1, 3, 5, 7} are reserved for one of the two groups B or G, and the positions {2, 4,
6, 8} are reserved for the other group. Arrangements in each position set can be done in 4P4 ways.
Hence, the total number of arrangements is 2! �� 4P4 �� 4P4 = 2 �� 24 �� 24 = 1152. The answer is (B).
Mathematically, this problem is the same as the previous one. Just the expression (wording) of the problem
is different.
40. The University of Maryland, University of Vermont, and Emory University have each 4 soccer
players. If a team of 9 is to be formed with an equal number of players from each university, how
many number of ways can the selections be done?
(A) 3
(B) 4
(C) 12
(D) 16
(E) 25
Selecting 3 of 4 players can be done in 4C3 =
4!
3!��1!
= 4 ways.
The selection from the 3 universities can be done in 3 �� 4 = 12 ways.
The answer is (C).
382 GMAT Math Bible
41. In how many ways can 5 persons be seated around a circular table?
(A) 5
(B) 24
(C) 25
(D) 30
(E) 120
For a circular table, we use cyclical permutations, not linear permutations. Hence, 1 in every r linear
permutations (here n = 5 and r = 5) is a circular permutation. There are 5P5 linear permutations. Hence,
5P5
5
=
5!
5
= 4 �� 3 �� 2 �� 1 = 24 permutations
The answer is (B).
42. In how many ways can 5 people from a group of 6 people be seated around a circular table?
(A) 56
(B) 80
(C) 100
(D) 120
(E) 144
For a circular table, we use cyclical permutations, not linear permutations. Hence, 1 in every r linear
permutations (here r = 5) is a circular permutation. There are 6P5 linear permutations and therefore 6P5
5
circular permutations. Now,
6P5
5
=
6!
5
= 1 �� 2 �� 3 �� 4 �� 6 = 144 permutations
The answer is (E).
43. What is the probability that a word formed by randomly rearranging the letters of the word ALGAE is
the word ALGAE itself?
(A) 1/120
(B) 1/60
(C) 2/7
(D) 2/5
(E) 1/30
The number of words that can be formed from the word ALGAE is
5!
2!
(A repeats). ALGAE is just one of
the words. Hence, the probability is
1
5!
2!
=
2!
5!
=
2
120
=
1
60
. The answer is (B).
of the words. Hence, the probability is
1
5!
2!
=
2!
5!
=
2
120
=
1
60
. The answer is (B).
