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Counting - 3

COMP109 Lectures

Combination Examples

The number of size 1 subsets of $\{1,2,3,4,5\}$ is 5. This is also the number of size 4 subsets of the set.

Any opposite subset sizes sample complements of each other and are therefore the same size.

Twelve people, including Mary and Peter, are candidates to serve on a committee of five. How many different committees are possible?

There are:

\[C(12,5)=\frac{12!}{(12-5)!5!}=792\]

This is as a result of choosing combinations of 5 people form a set of 12 people.

Of These How Many

  1. Contain both Mary and Peter? \(C(10,3)=120\) In this case we disregard them and adjust the samples and sample size accordingly.
  2. Contain neither Mary and Peter? \(C(10,5)=252\) In this case we select 5 members from the set without them.

  3. Contain either Mary or Peter (but not both)? \(2\times C(10,4)=420\)

    This is because we have two possible cases for Mary and Peter which have combinations $C(10,4)=210$.