Cambridge Additional Mathematics

Counting and the binomial expansion (Chapter 10) 267

Acombinationis a selection of objectswithoutregard to order.

For example, the possible teams of 3 people that can be selected from A, B, C, D, and E are:

ABC

BCD

CDE

ABD

BCE

ABE

BDE

ACD ACE ADE

There are 10 combinations in total.

Now given the five people A, B, C, D, and E, we know that there are 5 £ 4 £3=60permutations for

taking three of them at a time. So why is this 6 times larger than the number of combinations?

The answer is that for the combinations, order is not important. Selecting A, B, and C for the team is the

same as selecting B, C, and A. For each of the 10 possible combinations, there are 3! = 6 ways of ordering

the members of the team.

In general, when choosingrobjects fromnobjects,

number of combinations=number of permutations¥r!

=

n!

(n¡r)!

¥r!

=

n!

r!(n¡r)!

This is the binomial coefficient we encountered inSection C.

The number ofcombinationsonndistinct symbols takenrat a time is

¡n

r

¢

=

n!

r!(n¡r)!

Example 13 Self Tutor

How many different teams of 4 can be selected from a squad of 7 if:

a there are no restrictions b the teams must include the captain?

a There are 7 players up for selection and we want any 4 of them.

There are

¡ 7

4

¢

=35possible combinations.

b The captain must be includedandwe need any 3 of the other 6.

There are

¡ 1

1

¢

£

¡ 6

3

¢

=20possible combinations.

E Combinations

2 Copy and complete: Number of symbols Permutations in a line Permutations in a circle

1

2

3 6=3! 2=2!

4

3 If there arensymbols to be arranged around a circle, how many different cyclic permutations are

possible?

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_10\267CamAdd_10.cdr Monday, 23 December 2013 4:32:01 PM BRIAN