So the question has three parts and I'm stuck on the third, that's how it goes :

Claude wants to invite 5 friends to dinner in a pool of 11 friends.

a) How many ways can he do this?

My answer : $\binom{11}{5} = 462$

It's correct

b) 2 of these friends can't be separated from each other, how many ways?

My answer : $\binom{10}{4} = 210$

It's correct

c) 2 of them are mad at each other and can't be invited together, how many ways?

My answer : ''a) - b)'' = $\binom{11}{5} - \binom{10}{4} = 252$

It's false

The answer: 378

I can't seem to understand, plus there is no explanation in my textbook, it's straight the answer, so I'm completely lost. I've tried lots of different things but never got it right.

Thank you for helping me understand. Combinatorics are like hell to me.


Person's $A$ and $B$ cannot both be invited. There are $3$ possibilities.

Person $A$ comes and there $9C4$ ways to choose the other people. ($126$)

Person $B$ comes and there $9C4$ ways to choose the other people. ($126$)

Neither $A$ or $B$ comes and there $9C5$ ways to choose the other people. ($126$)

So there are $126+126+126=\color{blue}{378}$ ways to choose the people.

  • $\begingroup$ Omg you're a genius. Thank you I'm enlightened ! $\endgroup$ – Leorio Nov 26 '17 at 18:12

If you invite both, you can only choose $3$ more, from $9$,

so a slight modification of your effort yields $\binom{11}5 - \binom93 = 378$


Your question can be formulated using boolean algebra.

A:A is invited

B:B is invited

Your case can be formulated as

(A && ~B) || (~A && B) || (~A && ~B)

which is equivalent to

~(A && B)

So if you find number of ways in which both A and B are invited and subtract it from the total number of ways of inviting friends, you will have your answer

Total ways of inviting both A and B is $\binom{9}{3}$.

Total ways of inviting all $\binom{11}{5}$.

Hence your answer is $\binom{11}{5}$ - $\binom{9}{3}$.


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