Difference between revisions of "2005 AIME II Problems/Problem 1"
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== Problem == | == Problem == | ||
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A game uses a deck of <math> n </math> different cards, where <math> n </math> is an integer and <math> n \geq 6. </math> 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 <math> n. </math> | A game uses a deck of <math> n </math> different cards, where <math> n </math> is an integer and <math> n \geq 6. </math> 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 <math> n. </math> | ||
== Solution == | == Solution == | ||
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The number of ways to draw six cards from <math>n</math> is given by the [[binomial coefficient]] <math>{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}</math>. The number of ways to choose three cards from <math>n</math> is <math>{n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}</math>. We are given that <math>{n\choose 6} = 6 {n \choose 3}</math>, so <math>\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}</math>. Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 6\cdot6\cdot5\cdot4</math>. We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. With a little work we realize the factorization <math>8 \cdot 9 \cdot 10</math>, so <math>n - 3 = 10</math> and <math>n = 13</math> | The number of ways to draw six cards from <math>n</math> is given by the [[binomial coefficient]] <math>{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}</math>. The number of ways to choose three cards from <math>n</math> is <math>{n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}</math>. We are given that <math>{n\choose 6} = 6 {n \choose 3}</math>, so <math>\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}</math>. Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 6\cdot6\cdot5\cdot4</math>. We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. With a little work we realize the factorization <math>8 \cdot 9 \cdot 10</math>, so <math>n - 3 = 10</math> and <math>n = 13</math> | ||
== See Also == | == See Also == | ||
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*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||
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| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 16:32, 7 September 2006
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. With a little work we realize the factorization 
, so 
and 
