Difference between revisions of "2005 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
− | 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>. | + | 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>. |
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+ | 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>. | ||
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+ | 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>. | ||
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+ | Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 6\cdot6\cdot5\cdot4</math>. | ||
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+ | We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. | ||
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+ | 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 == |
Revision as of 02:14, 6 November 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
.