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Difference between revisions of "1990 AHSME Problems/Problem 16"

(Created page with "== 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 <math>13</math> married...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
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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).
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A) Since there are <math>13</math> men, the number of handshakes between only men is <math>\frac{13 \cdot 12}{2}=78</math>.
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B) Since there are <math>13</math> men and <math>12</math> women (excluding each man's spouse), there are <math>13 \cdot 12 = 156</math> ways.
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Adding this up, we get <math>156+78=234</math>
 
<math>\fbox{C}</math>
 
<math>\fbox{C}</math>
  

Revision as of 17:31, 14 September 2015

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 $156+78=234$ $\fbox{C}$

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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