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

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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>.
 
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>.
+
Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 720</math>.
  
We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s.
+
We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. Since 720 is close to <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <math>n = \boxed{013}</math>.
 
 
With a little work we realize the factorization <math>8 \cdot 9 \cdot 10</math>, so <math>n - 3 = 10</math> and <math>n = \boxed{013}</math>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 04:42, 14 August 2011

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 $(n - 3)(n - 4)(n - 5) = 720$.

We must find a factorization of the left-hand side of this equation into three consecutive integers. Since 720 is close to $9^3=729$, we try 8, 9, and 10, which works, so $n - 3 = 10$ and $n = \boxed{013}$.

See Also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions
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