Cambridge Additional Mathematics

Counting and the binomial expansion (Chapter 10) 263

Suppose we want to find the number of different permutations on the symbols A, B, C, D, E, F, and G,

taken 3 at a time.

There are 3 positions to fill:

In the 1 st position, any of the 7 symbols could be used, so we have

7 options.

This leaves any of 6 symbols to go in the 2 nd position, and this leaves

any of 5 symbols to go in the 3 rd position.

So, the total number of permutations=7£ 6 £ 5 fproduct principleg

=

7 £ 6 £ 5 £ 4 £ 3 £ 2 £ 1

4 £ 3 £ 2 £ 1

=

7!

4!

or

7!

(7¡3)!

Example 10 Self Tutor

A chess association runs a tournament with 16 teams. In how many different ways could the top

5 positions be filled on the competition ladder?

Any of the 16 teams could fill the ‘top’ position.

Any of the remaining 15 teams could fill the 2 nd position.

Any of the remaining 14 teams could fill the 3 rd position.

..

.

Any of the remaining 12 teams could fill the 5 th position.

The total number of permutations=16£ 15 £ 14 £ 13 £ 12

=16!

11!

= 524 160

1 st 2 nd 3 rd

1 st 2 nd 3 rd

7

1 st 2 nd 3 rd

7 6 5

16 15 14 13 12

1_st 2 nd_3_4rd th_5_th

If we are finding permutations on the complete set ofnsymbols, as inExample 9, then r=n, and the

number of permutations isn!.

The number ofpermutationsonndistinct symbols takenrat a time is:

n£(n¡1)£(n¡2)£::::£(n¡r+1)

| {z }

rof these

=

n!

(n¡r)!

So the top 5 positions could be filled in524 160ways.

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_10\263CamAdd_10.cdr Thursday, 16 January 2014 5:51:20 PM BRIAN