Cambridge Additional Mathematics

(singke) #1
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
cyan magenta yellow black Additional Mathematics

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100
Y:\HAESE\CAM4037\CamAdd_10\263CamAdd_10.cdr Thursday, 16 January 2014 5:51:20 PM BRIAN

Free download pdf