Difference between revisions of "2005 AIME II Problems/Problem 1"
m (→Solution) |
|||
Line 11: | Line 11: | ||
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) = | + | 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>. |
− | |||
− | |||
== See Also == | == See Also == |
Revision as of 04:42, 14 August 2011
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. Since 720 is close to , we try 8, 9, and 10, which works, so and .
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 |