1993 AIME Problems/Problem 7
Contents
Problem
Three numbers, , are drawn randomly and without replacement from the set
. Three other numbers,
, are then drawn randomly and without replacement from the remaining set of
numbers. Let
be the probability that, after suitable rotation, a brick of dimensions
can be enclosed in a box of dimension
, with the sides of the brick parallel to the sides of the box. If
is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Solution 1 (Combination)
Call the six numbers selected . Clearly,
must be a dimension of the box, and
must be a dimension of the brick.
- If
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
possibilities.
- If
is not a dimension of the box but
is, then both remaining dimensions will work as a dimension of the box. That gives us
possibilities.
- If
is a dimension of the box but
aren’t, there are no possibilities (same for
).
The total number of arrangements is ; therefore,
, and the answer is
.
Note that the in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers
whether they may be
or
.
Solution 2 (Hook Length Formula)
Like Solution 1, call the six numbers selected . Using the Hook Length Formula, the number of valid configuration is
. We proceed as Solution 1 does.
Solution 3 (Hook Length Formula)
As in the preceding solutions, we let where each
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
th Catalan number (where
is the number of pairs of rising and falling sections) is
Thus, there are
ways to pick which of
and
are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are
total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is
.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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