Jess invited all 8 of her friends to a party, but only 5 people showed up.

Question

Jess invited all 8 of her friends to a party, but only 5 people showed up. Given that two of her friends, Tom and Rick, refuse to be seen together, how many different possibilities are there for which five people showed up to her party?

I first answered the question by the Difference Rule:

All the ways of inviting 8 people and 5 of them showed up - the number of ways Tom and Rick come together = $\binom{8}{5}-\binom{6}{3}$

The correct answer I was given is $\binom{6}{5}+2\binom{7}{4}$ which applies the Addition Rule.

I understand how the correct answer comes but I don't know where my answer goes wrong. Can someone give some hints? Thank you!

Answer 1

The answer of $\binom 65 + 2 \binom 74$ should have instead been $\binom65 + 2\binom 64$. There are $6$ people other than Tom and Rick: we either choose $5$ of them, in $\binom 65$ ways, or choose $4$ of them plus one of Tom or Rick, in $2 \binom 64$ ways.

Once we fix the mistake, this gives the same result as your approach.

Answer 2

If I understand the question right, then your answer is correct, and the "correct answer" is incorrect.

In the "correct answer" they presumably calculated $\binom{6}{5}$ ways with neither of Tom and Rick turning up, and $\binom{7}{4}$ is the number of ways one of them could turn up. The mistake is: we are looking at the number of ways exactly one of them would turn up. In other words, if (say) Tom turns up, the other four people come from the set of $6$ (rather than $7$) people, as none of them will be either Tom or Rick.

So the corrected "correct answer" would be $\binom{6}{5}+2\binom{6}{4}=6+2\cdot 15=36$, which is the same as your answer $\binom{8}{5}-\binom{6}{3}=56-20=36$.