Pigeonhole Principle - Pairs of Sets with distinct sums (a) Each of 15 men and 15 woman is given an integer between 1 and 100 inclusive with no integer appearing more than once. The value of a man-woman pair is the sum of their numbers. Prove that there are two man-woman pairs with the same value. 

(b) Suppose there are 14 men and 14 women as above. Prove that there are still two man-woman pairs with the same value.

(c) Give an example with 13 men and 13 women where the value of every man-woman pair is unique. 

I can prove (a), namely that the set of possible sums is $S$ = {$3,4,5,...,197,198,199$} with |$S$| = $197$. With 15 men and women there are $15\cdot 15 = 225$ pairs so by the pigeonhole principle there will be at least two pairs with the same sums because $225&gt;197$.

With 14 men and women there are $14\cdot 14 = 196$ possible pairs but this is not larger than |$S$| = $197$. I assume that due to the way this is set up it is impossible to have certain pairs together so the actual amount of sums you could have is less than 197 but I don't know how to prove this.

For (c) I really have no clue how to come up with the lists without brute-force which isn't really working for me.

Edit: A basically identical problem has already been asked and solved here: Pigeon-Hole Principle Common Sum
You cannot get all of {3,...,17}.  If you have both 3 and 4, they must be 1+2 and 1+3, so 2 and 3 are on the same side. Then to also get 5, 4 must be on the same side as 2 and 3. Repeat this argument up to 16.  But then you cannot get 17.  A similar argument shows that you cannot get all 15 of {185,....,199}.
You cannot get all of {3,...,17}.  If you have both 3 and 4, they must be 1+2 and 1+3, so 2 and 3 are on the same side. Then to also get 5, 4 must be on the same side as 2 and 3. Repeat this argument up to 16.  But then you cannot get 17.  A similar argument shows that you cannot get all 15 of {185,....,199}.