1993 AIME Problems/Problem 7
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 997 numbers. Let be the probability that, after a suitable rotation, a brick of dimensions can be enclosed in a box of dimensions , 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
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 .
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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