Difference between revisions of "1993 AIME Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Three numbers, <math>a_1\,</math>, <math>a_2\,</math>, <math>a_3\,</math>, are drawn randomly and without replacement from the set <math>\{1, 2, 3, \dots, 1000\}\,</math>. Three other numbers, <math>b_1\,</math>, <math>b_2\,</math>, <math>b_3\,</math>, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let <math>p\,</math> be the probability that, after a suitable rotation, a brick of dimensions <math>a_1 \times a_2 \times a_3\,</math> can be enclosed in a box of dimensions <math>b_1 \times b_2 \times b_3\,</math>, with the sides of the brick parallel to the sides of the box. If <math>p\,</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | + | Three numbers, <math>a_1\,</math>, <math>a_2\,</math>, <math>a_3\,</math>, are drawn randomly and without replacement from the [[set]] <math>\{1, 2, 3, \dots, 1000\}\,</math>. Three other numbers, <math>b_1\,</math>, <math>b_2\,</math>, <math>b_3\,</math>, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let <math>p\,</math> be the [[probability]] that, after a suitable rotation, a brick of dimensions <math>a_1 \times a_2 \times a_3\,</math> can be enclosed in a box of dimensions <math>b_1 \times b_2 \times b_3\,</math>, with the sides of the brick [[parallel]] to the sides of the box. If <math>p\,</math> is written as a [[fraction]] in lowest terms, what is the sum of the numerator and denominator? |
== Solution == | == Solution == | ||
− | {{ | + | Call the six numbers selected <math>\displaystyle 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>b_6</math> must be a dimension of the brick. |
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+ | *If <math>x_2</math> 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 <math>3</math> possibilities. | ||
+ | *If <math>x_2</math> is not a dimension of the box but <math>x_3</math> is, then both remaining dimensions will work as a dimension of the box. That gives us <math>2</math> possibilities. | ||
+ | *If <math>x_4</math> is a dimension of the box but <math>x_2,\ x_3</math> aren’t, there are no possibilities (same for <math>x_5</math>). | ||
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+ | The total number of arrangements is <math>{6\choose3} = 20</math>; therefore, <math>p = \frac{3 + 2}{20} = \frac{1}{4}</math>, and the answer is <math>1 + 4 = 005</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=6|num-a=8}} | {{AIME box|year=1993|num-b=6|num-a=8}} | ||
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+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 19:40, 27 March 2007
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 .
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 |