Difference between revisions of "2005 AIME II Problems/Problem 1"
Line 19: | Line 19: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 22:22, 4 July 2013
Problem
A game uses a deck of different cards, where is an integer and The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find
Solution
The number of ways to draw six cards from is given by the binomial coefficient .
The number of ways to choose three cards from is .
We are given that , so .
Cancelling like terms, we get .
We must find a factorization of the left-hand side of this equation into three consecutive integers. Since 720 is close to , we try 8, 9, and 10, which works, so and .
See Also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.