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

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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>
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== Video Solution ==
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https://youtu.be/IRyWOZQMTV8?t=150
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~ pi_is_3.14
  
 
== Solution ==
 
== Solution ==
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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>.
  
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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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Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 720</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>.
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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{13}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AIME  box|year=2005|n=II|before=First Question|num-a=2}}
 
{{AIME  box|year=2005|n=II|before=First Question|num-a=2}}
  
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[[Category:Introductory Combinatorics Problems]]
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[[Category:Introductory Combinatorics Problems, Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:54, 31 July 2023

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 $n.$

Video Solution

https://youtu.be/IRyWOZQMTV8?t=150

~ pi_is_3.14

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 $n = \boxed{13}$.

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

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