1990 AHSME Problems/Problem 16
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[hide]Problem
At one of George Washington's parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If married couples attended, how many handshakes were there among these people?
Solution
We split this problem into two cases: A) The number of ways that men can shake hands with other men B) The number of ways that the men can shake hands with the other women (excluding their spouse).
A) Since there are men, the number of handshakes between only men is .
B) Since there are men and women (excluding each man's spouse), there are ways.
Adding this up, we get , so the answer is
Solution 2
We approach this by complement theory.
As there are 26 people, total handshakes between then are (26.25)/2=325
Now we exclude the cases when A)handshakes of only women takes place ,
and b) when husbands do handshakes with their spouse.
A)(13.12)/2 =78
B)13
subtracting these two cases from 325, we get 234, so the answer is 234
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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