GRE_Math_Bible_eBook

Permutations & Combinations 441

23.In how many ways can the letters of the word ACCLAIM be rearranged such that the vowels always appear together?

24. In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s appear together?

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?

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?

442GRE 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!

1

(E) 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?

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?

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?

Permutations & Combinations 443

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)6!5! 7!

(C)5!3! 7!

(D)3!5! 7!

(E)7!5! 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

444GRE 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 hav e 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

Permutations & Combinations 445

Answers and Solutions to Problem Set AA

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).

446GRE 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 =3!3P2 = (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 447

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

910 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).

448GRE 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

n!

3C2 = r!(n r)! =

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

The answer is (B).

Permutations & Combinations 449

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

4P4

(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

450GRE 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?

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

5 P5 =

2!

5! 0! 2! =

5 P3

The answer is (B).