Difference between revisions of "1950 AHSME Problems/Problem 45"
Mrdavid445 (talk | contribs) (Created page with "==Problem== The number of diagonals that can be drawn in a polygon of 100 sides is: <math>\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(...") |
(Added a solution to the problem. Lots of these problems don't even have solution sections.) |
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\textbf{(D)}\ 98 \qquad | \textbf{(D)}\ 98 \qquad | ||
\textbf{(E)}\ 8800</math> | \textbf{(E)}\ 8800</math> | ||
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| + | == Solution == | ||
| + | Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be <math>\binom{100}{2}=4950</math>. However this also counts the 100 sides of the polygon, so the actual answer is <math>4950-100=\boxed{\textbf{(A)}\ 4850 }</math>. | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 50p box|year=1950|num-b=44|num-a=46}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 11:20, 23 April 2012
Problem
The number of diagonals that can be drawn in a polygon of 100 sides is:
Solution
Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be 
. However this also counts the 100 sides of the polygon, so the actual answer is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 44 |
Followed by Problem 46 | |
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| All AHSME Problems and Solutions | ||
