counting

Question

A person has 8 friends, of whom 5 will be invited to a party.How many choices if 2 of the friends will only attend together??

Comments

is it $\begin{align*}\binom{6}{5} + \binom{6}{3} \end{align*}$ ??

 or, $\begin{align*} \binom{8}{6}*\binom{6}{5} + \binom{8}{2}*\binom{6}{3} \end{align*}$ ??

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1st one is right .

Answer 1

There will be two cases here ( Basically it is a problem of inclusion and exclusion principle

1. when both of them are coming :

So we have 8 people out of which 2 have already been chosen so our choice reduces to choosing 6 people out of 3 people so 6C3

2. When both of them are not coming :

So we have to  choose 5 people out of 6 people so 6c5

Hence the total number of ways are 6C3 + 6C5