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 Multi-Birth Convention (Posted on 2012-01-05)
At a convention of pairs of twins and sets of triplets, there were 9 sets of twins and 6 sets of triplets, each set being from a different family. Each twin shook hands with all the twins except his or her sibling and with half the triplets. Each triplet shook hands with all the other triplets except his or her siblings and with half the twins.

How many handshakes took place all together?

 See The Solution Submitted by Charlie Rating: 4.0000 (2 votes)

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 I think this works. Comment 1 of 1
So there are 18 people who have a twin, and 18 who have a triplet. .

Each twin shakes 16 other twins and 9 of the triplets.  This is 25 handshakes per twin.  Times 18 twins gives 450 handshakes given by twins.
Each trip shakes 15 other trips and 9 of the twins.  This is 24 handshakes per trip.  Times 18 gives 441 handshakes given by triplets.
450+432 = 882
Finally divide this by two because each handshake has been counted twice.
882/2 = 441 total.

I used the same reasoning with a simplified version: only 3 sets of twins and 2 sets of triplets which gives 39 handshakes.  I was able to check this smaller solution with a diagram.

 Posted by Jer on 2012-01-05 12:30:13

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