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Permutations & Combinations 431

Permutations (repetition allowed) using Indistinguishable Objects

By replacing C with A in the base set {A, B, C}, we get {A, B, A}. Reducing the repetitive (redundant) permutations yields

Table III

The Permutations (allowing Repetitions) are as follows [n = 3, r = 3], and two of the three objects are indistinguishable. The table is derived by replacing C with A in Table I and eliminating the repeating entries. Shaded entries are redundant and therefore not counted. (That is, we pick only one of the indistinguishable permutations.)

Total number of ways: 8

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Permutations (repetition not allowed) with Indistinguishable Objects

Indistinguishable objects are items that repeat in the original set. For example, replace C in the above set with A. Then the new base set would be {A, B, A}. Hence, if we replace C with A in the Table II, we get the repetitions in the permutations. Reducing the repetitive permutations yields

Table IV

The Permutations (not allowing Repetitions) with Indistinguishable objects are as follows [n = 3, r = 3]. The table is derived by replacing C with A in Table II and eliminating the repeating entries. Shaded entries are redundant and therefore not counted. (That is, we pick only one of the indistinguishable permutations)

Total number of ways: 3

So far, we have discussed the types of the problems. When trying to solve a problem, it is very helpful to identify its type. Once this is done, we need to count the number of possibilities.

Distinction between Indistinguishable Objects Problems and Repetition Problems

Suppose you are to arrange the letters of the word SUCCESS.

The base set is {S, U, C, C, E, S, S}. There are 3 S’s, which are indistinguishable objects. Hence, the letter S, can be used a maximum of 3 times in forming a new word if repetition is not allowed. So, SSSSUCE is not a possible arrangement.

If repetition is allowed, you can use S as many times as you wish, regardless of the number of S’s in the base word (for example, even if there is only 1 S, you can use it up to maximum allowed times). Hence, SSSSSSS is a possible arrangement.

Counting

There are three models of counting we can use.

We already discussed that if there are m combinations possible from a base set and if there are n permutations possible for each combination, then the total number of permutations possible is m n.

This is also clear from the Fundamental Principle of Counting.

Model 1:

The Fundamental Principle of Counting:

Construct a tree diagram (we used tables above) to keep track of all possibilities. Each decision made produces a new branch. Finally, count all the allowed possibilities.

The previous tables are examples of tree diagrams. They can represent possibilities as trees. The possibilities are also counted in the tables.

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Model 2:

Divide a work into mutually independent jobs and multiply the number of ways of doing each job to find the total number of ways the work can be done. For example, if you are to position three letters in 3 slots, you can divide the work into the jobs as

1)Choose one of three letters A, B, and C for the first position

2)Choose one of the remaining 2 letters for the second position

3)Choose the only remaining letter for the third position

This can be done in 3 2 1 = 6 ways. The model is a result of the Fundamental Principle of Counting.

Model 3:

Models 1 and 2 are fundamental. Model 3 uses at least one of the first two models. Here, we use the following formula:

Total Number of Permutations = Number of Combinations Number of Permutations of Each Combination

Predominantly, we use the model for calculating combinations. The total number of permutations and the number of permutations for each combination can be calculated using either or both models 1 and 2 in many cases.

Cyclic Permutations

A cyclic permutation is a permutation that shifts all elements of a given ordered set by a certain amount, with the elements that are shifted off the end inserted back at the beginning in the same order, i.e., cyclically. In other words, a rotation.

For example, {A, B, C, D}, {B, C, D, A}, {C, D, A, B}, and {D, A, B, C} are different linear permutations but the same cyclic permutation. The permutations when arranged in cyclic order, starting from, say, A and moving clockwise, yield the same arrangement {A, B, C, D}. The following figure helps visualize this.

Cyclic arrangements of the cyclic permutations.

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For the r placement positions (for the example in the figure, r equals 4), we get r permutations, each is an equivalent cyclic permutation. Hence, the number of cyclic permutations equals

(The number of ordinary permutations) ÷ r

Hence, for nPr permutations, nPr ÷ r cyclic permutations exist. Simply said, r linear permutations would be same cyclic permutation.

Also, {A, B, C, D} and {A, C, B, D} are different linear permutations and different cyclic permutations, because arranging them in cyclic order yields different sequences.

Factorial

The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. That is, n! = n(n – 1)(n – 2) 3 2 1. For example, 4! = 4 3 2 1 = 24. Note: 0! is defined to be 1.

Formulas

Verify that the following formulas apply to the scenarios mentioned above. These formulas should be memorized.

Formula 1: If you have n items to choose from and you choose r of them, then the number of permutations with repetitions allowed is

n n . . . n = nr

(r times)

Formula 2: The formula for permutations with repetitions not allowed is

n!

nPr = (n r)!

Formula 3: The formula for combinations with repetitions not allowed is

n! nCr = r!(n r)!

Formula 4: We know that k distinguishable objects have k! different arrangements (permutations). But a set of k indistinguishable objects, will have only 1 indistinguishable permutation. Hence, if we have P

The case is similar when we have more than one set of indistinguishable objects. Suppose the word

n Pr

3! 2! permutations because here we have a set of 3 indistinguishable objects A and a set of 2 indistinguishable

objects B.

There are formulas for the other problem models, but they are not needed for the test. We can always use the Fundamental Principal of Counting for them.

Formula 5: For r linear positions (for the example in the figure, r equals 4), we get r permutations, each of which is an equivalent cyclic permutation. Hence, the number of cyclic permutations is

(The number of ordinary permutations) ÷ r

The formulas in this section will be referenced while we solve the problems.

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Problem Solving Strategy

In permutation and combination problems, it is very important to recognize the type of problem. Many students mistakenly approach a combination problem as a permutation, and vice versa. The steps below will help you determine the problem type.

Solving a permutation or combination problem involves two steps:

1)Recognizing the problem type: permutation vs. combination.

2)Using formulas or models to count the possibilities.

We have three questions to ask ourselves in order to identify the problem type:

1)Is it a permutation or combination?

Check any two typical arrangements with the same combination. If the two arrangements are counted only once, it is a combination problem. Otherwise, it is a permutation.

For example, if you are asked for a lock code, then 321 and 123 could be two possibilities, and the two numbers are formed from the same combination (Same number of 1’s, 2’s, and 3’s). So, lock codes must be permutations.

For another example, suppose you have 5 balls numbering 1 through 5. If you are asked to select 3 out of the 5 balls and you are only interested in the numbers on the balls, not the order in which they are taken, then you have a combination problem.

Problems that by definition connote ordering (though not directly stated) are permutations. For example, 3 digits form a 3-digit number. Here, the 3-digit number connotes ordering. For another example, if you are to answer 3 questions, you probably would not be asked to answer a particular question more than once. So, you would not allow repetition in the calculations. Though not often needed, such logical assumptions are allowed and sometimes expected.

If the problem itself defines slots for the arrangements, it is a permutation problem. Words like “arrange” define slots for the arrangements. We will explain this in more detail later in the problems.

Generally, “arrangements” refer to permutations, and “selections” refer to combinations. These words often flag the problem type.

Other words indicating permutations are “alteration,” “shift,” “transformation,” and “transmutation,” all of which connote ordering.

For example:

In how many ways can the letters of the word XYZ be transformed to form new words? In how many ways can the letters of the word XYZ be altered to form new words?

Some words indicating combinations are “aggregation,” “alliance,” “association,” “coalition,” “composition,” “confederation,” “gang,” “league,” and “union,” (all of which have nothing to do with arrangements but instead connote selections.)

For example:

In how many ways can a coalition of 2 countries be formed from 4 countries?

(Here, a coalition is the same whether you say country A and B are a coalition or country B and country A are a coalition.)

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2)Are repetitions allowed?

Check whether, based on the problem description, the results of a permutation/combination can have repetitions.

For example:

If you are to list countries in a coalition, you can hardly list a country twice. (Here, repetition automatically is not allowed unless specified otherwise.)

If you have 3 doors to a room, you could use the same door for both entering and exiting. (Here, repetition is automatically allowed.)

3)Are there any indistinguishable objects in the base set?

Check the base set: the objects from which a permutation or a combination are drawn. If any indistinguishable objects (repetitions at base set level) are available, collect them. This is easy since it only requires finding identical objects in a base set, which is usually given.

For example, if the original question is to find the words formed from the word GARGUNTUNG, then, in this step, you collect the information: G exists thrice, U exists twice, and so on.

Once the problem type is recognized, use the corresponding formula or model to solve it.

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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

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

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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

4 C4

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?

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?

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

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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?

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

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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

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

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

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!