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Difference between revisions of "1993 AIME Problems/Problem 7"

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== Solution 2 ==
 
== Solution 2 ==
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Like solution 1, call the six numbers selected <math>x_1 > x_2 > x_3 > x_4 > x_5 > x_6</math>. Clearly, <math>x_1</math> must be a dimension of the box, and <math>x_6</math> must be a dimension of the brick. Using the hook-length formula, the number of valid configuration is <math>\frac{6!}{4\cdot3\cdot2\cdot3\cdot2}</math>. This gives us 5, and we proceed as solution 1 did.
+
Like solution 1, call the six numbers selected <math>x_1 > x_2 > x_3 > x_4 > x_5 > x_6</math>. Using the hook-length formula, the number of valid configuration is <math>\frac{6!}{4\cdot3\cdot2\cdot3\cdot2}</math>. This gives us <math>5</math>, and we proceed as solution 1 did.
  
 
== See also ==
 
== See also ==

Revision as of 20:04, 27 October 2015

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.

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
All AIME Problems and Solutions

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