I'm revising for finals and I have come across this following question:

a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.

b) Repeat a) but 2 of the friends are married and will not attend separately.

For a) I got ${11 \choose 5} = \frac{11!}{(11-5)! \cdot 5!} = 462$.

I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then: ${9 \choose 5} = \frac{9!}{(9-5)! \cdot 5!} = 126$, but I'm pretty sure this is wrong,

Could someone explain b) to me?

Thanks in advance!


Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.

  • $\begingroup$ 10!(5!x5!)=252 462-252=210. I get it Thanks! :) $\endgroup$ – Xabi Apr 22 '12 at 15:07

for part b,

the woman has 2 choices , either to invite the couple or not invite them ,

CASE1: she invites them

then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 \choose 3}$ ways

CASE2: when she doesnt invites the couple

then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 \choose 5}$ ways


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