Combinatorics Word Problem (Inclusion-Exclusion)

I'm just unsure if I'm interpreting this problem right:

At a 12-week conference in mathematics, Sharon met seven of her friends from college. During the conference she met each friend at lunch 35 times, every pair of them 16 times, every trio eight times, every foursome four times, each set of five twice, and each set of six one, but never all seven at once. If she had lunch every day during the 84 days of the conference, did she ever have lunch alone?

So I've been working this out in my head for a while and I'm fairly sure the following is correct, but can anyone verify? $$84 - 35 \binom{7}{1} + 16 \binom{7}{2} - 8 \binom{7}{3} + 4 \binom{7}{4} - 2 \binom{7}{5} + \binom{7}{6}=0$$

And this is basically taking the number of days of her conference and subtracting all the ones that she had lunch with people. Because we have a result of zero, she didn't have lunch alone at all.. Correct?

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Yes, it is correct. – Sasha Oct 14 '12 at 3:33

For the sake of having an accepted answer: Yes, it is correct.

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I get $5$. Doing the calculations you list above, which I agree with I get:

$84 - 245 + 341 - 280 + 140 - 42 + 7 = 5$

$5$ lunches eaten alone.

I know this is an old question, however if anyone comes across it in the same manner that I did - I'd like to see the correct outcome posted.

I actually did my problem a little differently in that I did this:

$245-341+280-140+42-7 = 79$ lunches shared with friends $84-79 = 5$ lunches alone

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I'm not going to go through all of the addition mentally, but just looking at the sum $\pmod {10}$ obtains that the last digit should be $0$. Check your work. – Tim Ratigan Dec 2 '13 at 7:46