1990 AHSME Problems/Problem 16

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 $13$ married couples attended, how many handshakes were there among these $26$ people?

$\text{(A) } 78\quad \text{(B) } 185\quad \text{(C) } 234\quad \text{(D) } 312\quad \text{(E) } 325$

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 $13$ men, the number of handshakes between only men is $\frac{13 \cdot 12}{2}=78$ .

B) Since there are $13$ men and $12$ women (excluding each man's spouse), there are $13 \cdot 12 = 156$ ways.

Adding this up, we get $156+78=234$ , so the answer is $\fbox{C}$

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

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1990 AHSME Problems/Problem 16

Contents

Problem

Solution

Solution 2

See also