Difference between revisions of "1990 AHSME Problems/Problem 16"
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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. | 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. | ||
| − | Adding this up, we get <math>156+78=234</math> | + | Adding this up, we get <math>156+78=234</math>, so the answer is |
<math>\fbox{C}</math> | <math>\fbox{C}</math> | ||
Revision as of 17:37, 24 February 2016
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
