Difference between revisions of "2005 AIME II Problems/Problem 1"

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== See Also ==
 
== See Also ==
  
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*[[2005 AIME II Problems/Problem 2| Next problem]]
 
*[[2005 AIME II Problems]]
 
*[[2005 AIME II Problems]]
  
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 20:53, 7 September 2006

Problem

A game uses a deck of $n$ different cards, where $n$ is an integer and $n \geq 6.$ 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 $n.$

Solution

The number of ways to draw six cards from $n$ is given by the binomial coefficient ${n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$. The number of ways to choose three cards from $n$ is ${n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$. We are given that ${n\choose 6} = 6 {n \choose 3}$, so $\frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 6 \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$. Cancelling like terms, we get $(n - 3)(n - 4)(n - 5) = 6\cdot6\cdot5\cdot4$. We must find a factorization of the left-hand side of this equation into three consecutive integers. With a little work we realize the factorization $8 \cdot 9 \cdot 10$, so $n - 3 = 10$ and $n = 13$

See Also