2005 AIME II Problem 1

Question

A game uses a deck of

n

different cards, where

n

is an integer and

n \ge 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.

Answer

Write the ratio

\frac{\binom{n}{6}}{\binom{n}{3}}

in factorials; almost everything cancels

Solve

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

by writing

720 = 10 \cdot 9 \cdot 8

The condition says

\binom{n}{6} = 6\binom{n}{3}.

Dividing the binomial coefficients,

\begin{aligned} \frac{\binom{n}{6}}{\binom{n}{3}} &= \frac{(n-3)(n-4)(n-5)}{6 \cdot 5 \cdot 4} \\ &= 6, \end{aligned}

so

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

= 10 \cdot 9 \cdot 8.

Since the product

(n-3)(n-4)(n-5)

is increasing in

n,

the only solution is

n - 3 = 10,

that is,

n = 13.

Problem 1 in other years