     # Section 13.2 – Permutations and Combinations .281 Section 13.2 – Permutations and Combinations

• View
219

0

Embed Size (px)

### Text of Section 13.2 – Permutations and Combinations .281 Section 13.2 – Permutations and Combinations

• 281

Section 13.2 Permutations and Combinations

Objective #1: Solving Counting Problems Using Permutations involving n Distinct Objects. A code for an alarm may consists of a four-digit number with digits ranging from 0 to 9. The counting principle tells us that there at 10101010 = 10,000 different possible codes. The order that the digits are typed in is important since the code 2271 is different from 1722 and would not work in place of 1722. Such an arrangement of digits is called a type of permutation where each object is distinct and repetition is allowed. In this instance, we have 10 distinct digits and repetition is allowed in selecting four digits to make the code. We will also see that there are two other types of permutations as well. Definition A permutation is an arrangement of r objects in a specific order chosen from n objects. With permutations, order is important (i.e., order makes a difference). Types of Permutations: 1) Distinct with Repetition: The n objects are distinct and repetition is allowed in the selection of r number of objects.

2) Distinct without Repetition: The n objects are distinct and repetition is not allowed in the selection of r number of objects.

3) Not Distinct: The n objects are not distinct and all of the objects are used in the arrangement. As illustrated by the alarm code example, we use the counting principle to solve the first type of permutation.

Solve the following: Ex.1 How many three-letter codes can be formed from the letters A, B, C, D, E, F, G, and H if repetition is allowed? Solution: Since we have 8 distinct letters and repetition is allowed in the selection of 3 letters, then the number of three-letter codes that can be formed is 888 = 83 = 512 codes. This leads to the following theorem.

• 282

Theorem: Permutations of Distinct Objects with Repetition The number of ordered arrangements of r objects selected from n distinct objects where repetition is allowed is nr. The second type of permutation is where we are selecting r number of objects from n distinct objects and repetition is not allowed. This implies that r n since we cannot select more objects than the number that exist. Solve the following: Ex.2 How many three-letter codes can be formed from the letters A, B, C, D, E, F, G, and H if repetition is not allowed? Solution: For the first letter, we have eight choices, but since repetition is not allowed, there are only seven letters to choose from for the second letter and only six letters to choose from for the last letter. Thus, the number of codes is 876 = 336 codes. In this example, we are asking how many ways can 8 letters be arranged in order using three letters that do not repeat. We will use the following notation to represent this question: P(8, 3). This means that we have a permutation of eight objects taken 3 at a time without repetition where order is important. Thus, P(8, 3) = 876 = 336. Ex. 3 How many ways can five boxes be stacked? Solution: Since the five boxes are distinct and we cannot repeat a box in the stack, then we have a permutation of 5 objects taken 5 at a time without repetition: P(5, 5) = 54321 = 120 There are 120 ways to stack the five boxes. To help us derive a formula for the second type of permutation, consider the following: P(8, 3) = 876 (multiply top & bottom by 54321) =

8765432154321

(use the definition of factorial)

=

8!5!

(rewrite 5 as (8 3))

=

8!(83)!

We can now generalize this result.

• 283

Theorem: Permutations of Distinct Objects without Repetition The number of ordered arrangements of r objects selected from n distinct objects where repetition is not allowed is: nPr = P(n, r) =

n!(nr)!

where r n

Evaluate the following: Ex. 4a P(9, 3) Ex. 4b P(4, 1) Ex. 4c P(83, 4) Solution: a) P(9, 3) =

9!(93)!

=

9!6!

=

9876!6!

= 987 = 504

b) P(4, 1) =

4!(41)!

=

4!3!

=

43!3!

= 4

c) P(83, 4) =

83!(834)!

=

83!79!

=

8382818079!79!

= 83828180

= 44,102,880 You can actually calculate the number directly on a scientific or graphing calculator using the nPr key. Thus, you would type 83 nPr 4 and hit equals/enter. To access the nPr key on a TI-30XA, you would type 2nd and then the 9 key. For a TI-84 plus, you would hit the Math key, arrow over to the PRB menu and hit 2: nPr. Solve the following: Ex. 5 In how many ways can four people each have a different birthday assume that there are 365 days in a year? Solution: We have a permutation of 365 days taken 4 at a time without repetition where order is important. P(365, 4) =

365!(3654)!

=

365!361!

=

365364363362361!361!

= 365364363362

= 17,458,601,160 There are 17,458,601,160 different ways four people could have different birthdays. Objective #2: Solving Counting Problems Using Combinations. Recall that in the alarm code example at the beginning of the section, the codes 1722 and 2271 are different permutations since the order of the digits make a difference. There are many instances where order does not make a difference. For instance, in blackjack, the order in how one receives the cards does not matter since it makes no difference if we get an ace first

• 284

and a ten second or vice-versa. What matters is that one card is an ace and the other is a ten. Such an arrangement of objects where order is not important is called a combination. Definition A combination is an arrangement of r objects chosen from n objects. where order of how the objects are selected does not matter and repetition is not allowed. With combinations, order is irrelevant (i.e., order makes no difference). The notation C(n, r) means that we have a combination of n objects taken r at a time without repetition where order is not important. In combinations, the arrangements ADS and DSA are consider to be the same combination since order does not matter. This is a much more formal definition of combinations than what we experience in the everyday world. When discuss the combination for a lock, it is not actually a combination, but a permutation since the order does matter for a lock. Solve the following: Ex. 6 List all the combinations of a, b, c, d, and e taken 3 at a time. What is C(5, 3)? Solution: Since we are looking for combinations, the arrangement abc and acb are considered to be the same. Thus, we just need to systematically list all the combinations: abc, abd, abe, acd, ace, ade, bcd, bce, bde, and cde Thus, we have ten possible combinations. This means the C(5, 3) = 10. In a permutation, we can order three letters 3! or 6 different ways. So, if order had been important in the last example, the total number of permutations would have been 3!C(5, 3) = 610 = 60 permutations. But, P(5, 3) =

5!(53)!

= 60 which implies that 3!C(5, 3) = P(5, 3) =

5!(53)!

. If we

solve for C(5, 3), we get: 3!C(5, 3) =

5!(53)!

(divide both sides by 3!)

C(5, 3) =

5!(53)!3!

We can then generalize this result:

• 285

Theorem: Combinations of Distinct Objects without Repetition The number of arrangements of r objects selected from n distinct objects where repetition is not allowed and order is not important is: nCr = C(n, r) =

n!(nr)!r!

where r n

Notice that this is essentially the same formula used for the binomial coefficient. Find the following: Ex. 7a C(8, 1) Ex. 7b C(7, 4) Ex. 7c C(n, n) Ex. 7d C(n, 0) Ex. 7e C(74, 71) Solution: a) C(8, 1) =

8!(81)!1!

=

87!7!1!

=

81

= 8

b) C(7, 4) =

7!(74)!4!

=

7654!3!4!

=

7656

= 35

c) C(n, n) =

n!(nn)!n!

=

n!0!n!

=

11

= 1

d) C(n, 0) =

n!(n0)!0!

=

n!n!0!

=

11

= 1

e) C(74, 71) =

74!(7471)!71!

=

74737271!3!71!

=

7473726

= 64,824

Ex. 8 How many committees of four people can be selected from nine people? Solution: Since the order that the people are selected is not important, then this is a combination problem without repetition. We will need to calculate C(9, 4) =

9!(94)!4!

=

98765!5!4!

=

98764!

=

302424

= 126

There are 126 ways the committee can be selected. Ex. 9 In how many ways can a committee of two teachers and three parents be selected from 8 teachers and 10 parents? Solution: We can first determine the number of ways the teachers can be selected and then the number of ways the parents can be selected. By the multiplication principle, the total number of ways the committee can be selected is the product of these two results. Teachers: C(8, 2) =

8!(82)!2!

=

876!6!2!

=

562

= 28

• 286

Parents: C(10, 3) =

10!(103)!3!

=

10987!7!3!

=

7206

= 120

Total: C(8, 2)C(10, 3) = 28120 = 3360 There are 3360 ways the committee can be selected. Objective #3: Solving Counting Problems Using Permutations of n Non-distinct Objects. We will now examine the third type of permutation, which was the Not Distinct case: n objects are not distinct and all of the objects are used in the arrangement. We will illustrate this with the next example. Solve the following: Ex. 10 How many ten letter words can you form from the word "basketball?" Solution: The word basketball has 2 b's, 2 l's, 2

Recommended ##### Permutations and Combinations. Objectives:  apply fundamental counting principle  compute permutations  compute combinations  distinguish permutations
Documents ##### BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS CHAPTER
Documents ##### Permutations and Combinations Standards: MM1D1b. Calculate and use simple permutations and combinations
Documents ##### PERMUTATIONS AND COMBINATIONS BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED FACTORIAL
Documents Documents