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# there are eleven divers on a swim team. if the team can only send four divers to a competition, how many ways can they send four?

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### James

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## 2B Permutations

2B  Permutations

Suppose you wish to take a picture of your three friends, Alice, Betty, and Cindy, lined up in a row. As photographer you can arrange the women from left to right in the order you prefer. In how many different ways can you line up the women? As there are only three women, it is not difficult to ponder a moment and write down all the possibilities; referring to the ladies by the letters A, B, C, we discover the six arrangements

ABC ACB BAC BCA CAB CBA

Each one of the arrangements of the women is called a ; altogether there are six possible permutations.

But what if there were ten ladies? Now the problem becomes more serious, as it would take forever to write down all the possibilities, and even if you tried you would never know whether you left any out. Thus it appears that we must devise a method of counting the permutations without actually writing them all down. The way to do this is to divide the procedure of lining up the women into steps, and then use the multiplication principle.

Going back to the case of three ladies, we imagine that there are three slots where the ladies may stand:

In arranging the ladies there are three corresponding steps; we must fill the first slot, then the second slot, and then the third. We have three ways to place a lady in the first slot. After this first step is complete, there are then two ways of filling the second slot, as there are only two ladies left. Finally, after filling the first two slots, there is only one lady left, so there is just one way to fill the last slot. By the multiplication principle, the number of ways of completing the entire procedure by doing all three steps is

3 · 2 · 1 = 6 .

Of course the same technique works with ten ladies, except that we need ten slots and ten steps instead of three. The number of permutations of ten women is

10! = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3,628,800 .

We use the shorthand notation 10! to refer to this long product. More generally, for any positive integer N the notation N! designates the product

N! = N · (N−1) · (N−2) · … · 3 · 2 · 1   .

We read "N!" as "N factorial". Some other factorials are

1! = 1 2! = 2 · 1 = 2 3! = 3 · 2 · 1 = 6

4! = 4 · 3 · 2 · 1 = 24

5! = 5 · 4 · 3 · 2 · 1 = 120

6! = 6 · 5 · 4 · 3 · 2 ·1 = 720 .

A of a group of objects is a way of placing the objects in an order. Since obviously our method of ordering ladies should work just as well with general objects, we have the following rule:

First Permutation Rule

The number of possible permutations of N objects is N!

There are many other ways of ordering objects besides lining them up for a picture, as the ensuing examples will demonstrate.

example 1

Six horses are entered in a race. In how many different orders can the horses complete the race?

There are six steps in ordering the horses. First we choose a first place winner, then a second place finisher, then third, fourth, fifth, and sixth. There are 6 choices for first, then 5 choices for second, 4 for third, etc.; therefore, the number of possible permutations is

6! = 6 · 5 · 4 · 3 · 2 ·1 = 720 .

example 2

A dance class has 5 men students and 5 women students. How many ways can the teacher form 5 couples to dance the waltz?

We divide the procedure of forming couples into 5 steps. First we number the women 1 through 5. There are 5 ways of choosing a male partner for woman # 1. Then for woman # 2 there are only 4 choices remaining, for woman # 3 only 3 choices, for woman # 4 there are 2 choices, and finally for woman # 5 there is but 1 choice. The number of ways the teacher can form 5 couples is

5! = 5 · 4 · 3 · 2 ·1 = 120.

example 3

A baseball team has 9 starting players. How many ways can the coach make out the batting order?

The coach may select the leadoff batter in 9 ways, then the second batter in 8 ways, the third batter in 7 ways, etc., until the last batter can be chosen in only one way. The end result is

9! = 9 · 8 · 7 · 6 · 5 ·4 · 3 · 2 · 1 = 362,880 .

example 4

Lori's dog gave birth to 4 puppies. As she has no space for more than one dog, she will give one puppy to each of her four cousins. In how many ways can she distribute the four puppies among her cousins?

The four steps are to give Fido to a cousin, in one of 4 ways, then Lassie to another cousin in one of 3 ways, then Rover to a cousin in one of 2 ways, and finally Spot to the last cousin. The number of possible permutations is

4! = 4 · 3 · 2 · 1 = 24 .

example 5

Four acrobats will line up for a picture. Each acrobat has the option of standing on his/her feet or standing on his/her head. How many possible pictures are there?

This problem combines our permutation formula with the multiplication principle. We can divide the procedure of lining up and positioning the acrobats into five steps:

Source : www.math.hawaii.edu

## 4. There are eleven divers on a swim team. If the

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## 4. There are eleven divers on a swim team. If the team can only send four divers to a competition, how many ways can they send four?

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4. There are eleven divers on a swim team. If the team can only send four divers to a competition, how many ways can they send four?

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## Bulletin of the Atomic Scientists

The Bulletin of the Atomic Scientists is the premier public resource on scientific and technological developments that impact global security. Founded by Manhattan Project Scientists, the Bulletin's iconic "Doomsday Clock" stimulates solutions for a safer world.

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## Bulletin of the Atomic Scientists

Ara 1970 104 sayfa 26. cilt,10. no. ISSN 0096-3402

Yayınlayan: Educational Foundation for Nuclear Science, Inc.

The Bulletin of the Atomic Scientists is the premier public resource on scientific and technological developments that impact global security. Founded by Manhattan Project Scientists, the Bulletin's iconic "Doomsday Clock" stimulates solutions for a safer world.

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