Question

A boy and his sister are included in a group of 8 boys and 6 girls. In how many ways can a tennis four( 2 boys and 2girls) be selected so as not to include both the boy and his sister?

Answer 1

385 ways

Explanation:

First off, we're working with combinations - we don't care in what order people are picked. The general formula for combinations is:

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

If we didn't have the restriction on the brother/sister, we'd have a combination with population of 8 boys choosing 2 multiplied by a combination with population of 6 girls choosing 2:

`(C_(8,2))(C_(6,2))=((8!)/((2!)(6!)))((6!)/((2!)(4!)))=(8!)/(2!2!4!)=40320/(2xx2xx24)=420 " ways."`

Since there is another restriction on our question, we know we can't have an answer that is higher than 420.

For our question, notice that if the brother is picked, the sister can't be. There are `(C_(8,2))=28` ways the boys can be picked. Of them, there are 7 combinations that involve the brother, and so there are 21 that don't. When the brother isn't picked, we can pick from all 6 girls. When the brother is picked, we can only pick from 5 girls (we exclude the sister). This gives:

`21xx(C_(6,2))+7xx(C_(5,2))=21xx15+7xx10=385`

Also note that this works the same way if we'd analyzed this from the point of view of the sister (15 different ways they can be picked) - if the sister is picked (5 ways), the brother can't be `(C_(7,2))` and if she isn't (10 ways) then he can be `(C_(8,2))`, giving:

`5xx(C_(7,2))+10xx(C_(8,2))=5xx21+10xx28=385`