Difference between revisions of "2007 AIME I Problems/Problem 10"
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[[Casework]] shows that: | [[Casework]] shows that: | ||
| − | *If column 1 and column 2 do not share any two filled squares on the same row, then there are <math> | + | *If column 1 and column 2 do not share any two filled squares on the same row, then there are <math>{6\choose3}</math> combinations. The same goes for columns 3 and 4, so we get <math>{6\choose3}^2 = 400</math>. |
*If column 1 and column 2 share 1 filled square on the same row, then there are <math>\ldots</math>. | *If column 1 and column 2 share 1 filled square on the same row, then there are <math>\ldots</math>. | ||
Revision as of 20:31, 14 March 2007
Problem
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let 
be the number of shadings with this property. Find the remainder when is divided by 1000.
Solution
Casework shows that:
- If column 1 and column 2 do not share any two filled squares on the same row, then there are
combinations. The same goes for columns 3 and 4, so we get 
. - If column 1 and column 2 share 1 filled square on the same row, then there are
.
combinations. The same goes for columns 3 and 4, so we get 
.
.Thus, there are 
number of shadings, and the solution is 
.
See also
| 2007 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
