# 1993 AIME Problems/Problem 7

## Problem

Three numbers, $a_1\,$, $a_2\,$, $a_3\,$, are drawn randomly and without replacement from the set $\{1, 2, 3, \dots, 1000\}\,$. Three other numbers, $b_1\,$, $b_2\,$, $b_3\,$, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let $p\,$ be the probability that, after a suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3\,$ can be enclosed in a box of dimensions $b_1 \times b_2 \times b_3\,$, with the sides of the brick parallel to the sides of the box. If $p\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

## Solution 1

Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$. Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick.

• If $x_2$ is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us $3$ possibilities.
• If $x_2$ is not a dimension of the box but $x_3$ is, then both remaining dimensions will work as a dimension of the box. That gives us $2$ possibilities.
• If $x_4$ is a dimension of the box but $x_2,\ x_3$ aren’t, there are no possibilities (same for $x_5$).

The total number of arrangements is ${6\choose3} = 20$; therefore, $p = \frac{3 + 2}{20} = \frac{1}{4}$, and the answer is $1 + 4 = \boxed{005}$.

Note that the $1000$ in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers $x_1,x_2,x_3,x_4,x_5,x_6$ whether they may be $1,2,3,4,5,6$ or $999,5,3,998,997,891$.

## Solution 2

Like solution $1$, call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$. Using the hook-length formula, the number of valid configuration is $\frac{6!}{4\cdot3\cdot2\cdot3\cdot2}$. This gives us $5$, and we proceed as solution 1 did.

## Solution 3

As in the preceding solutions, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected. It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter. We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain. The formula for the $n$th Catalan number (where $n$ is the number of pairs of rising and falling sections) is $$\frac{\binom{2n}{n}}{n+1}$$ Thus, there are $\frac{\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are $\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\frac{\left(\frac{\binom{6}{3}}{4}\right)}{\binom{6}{3}}=\frac14\Longrightarrow \boxed{005}$