GRE_Math_Bible_eBook

Permutations & Combinations 451

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

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).

452GRE 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

7 P7 =

2! 3!

6 5 4 3 2 1 = 2 6

5 4 3 =

60

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?

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 453

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).

Permutations & Combinations 455

19. In how many ways can 3 red marbles, 2 blue marbles, and 5 yellow marbles be placed in a row?

10!

(E) (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:

10! 3! 2! 5! = 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)6 P3 4 P1

10C4

(E)6 P3 4 P1

10 P4

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.

456GRE 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

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+6 C4

10P4

(D)6 P3 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 457

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

(A) 3! 3!

(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)2!5! 2!3!

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 52!P5 = 2!5! .

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 32!P3 = 2!3! .

458GRE Math Bible

Hence, by The Fundamental Principle of Counting, the total number of ways of rearranging the letters is

5! 3! 2! 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

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)7 P6

2! 2!

(C)7 P6 3 P3

2! 2!

(D)6 P6 3 P3

2! 2!

(E)10 P6 3 P3

2! 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 459

____ ____ _____ {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 62!P6 (E repeats twice) ways.

And the unit {S, S, M} can be internally arranged in 32!P3 ways.

Hence, by Model 2, the total number of ways the letters can be rearranged is

6 P6 3 P3

2! 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?

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!

1

(E) 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).

460GRE 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

24 1

(repetitions allowed). The probability is 25 = 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 23 23 23 23 = 1681. 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 23 23 23 23 = 1681 are the number of ways of doing the specific jobs, and the denominators are the number of ways of doing all possible jobs.