A probability problem by A Former Brilliant Member

Probability Level 2

Six couples were at a birthday party. If each person shook hands exactly once with everyone else except his/her spouse, how many handshakes were exchanged?

Note: One cannot shakehands with himself/herself.

60 66 56 48

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A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Richard Costen
Dec 27, 2016

There are 12 people so choose all pairs from the 12 to shake hands, ( 12 2 ) = 66 {12 \choose 2}=66 , and then subtract the 6 unwanted handshakes between spouses: 66 − 6 = 60 66-6=\boxed{60}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

There are a total of 12 12 people at the party. Each shakehands with everyone else except for their spouse which will be a total of 12 − 2 = 10 12-2=10 other people.The total number of handshakes will be ( 12 ) ( 10 / 2 ) = 60 (12)(10/2)=60 , where we divide by 2 2 to correct for counting each handshake twice.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

12 couples mean 24 people then

Nikkil V - 4 years, 5 months ago

It must be six couples. I edited the problem.

A Former Brilliant Member - 4 years, 5 months ago

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