Difference between revisions of "2007 AMC 10B Problems/Problem 20"
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The number of ways to choose <math>3</math> squares is <math>25 \cdot 16 \cdot 9,</math> but you must divide by <math>3!</math> since some sets are the same. | The number of ways to choose <math>3</math> squares is <math>25 \cdot 16 \cdot 9,</math> but you must divide by <math>3!</math> since some sets are the same. | ||
| − | <cmath>\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } | + | <cmath>\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } 600}</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2007|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2007|ab=B|num-b=19|num-a=21}} | ||
Revision as of 15:53, 1 November 2011
Problem
A set of 
square blocks is arranged into a 
square. How many different combinations of 
blocks can be selected from that set so that no two are in the same row or column?
Solution
There are ways to choose the first square. The four remaining squares in its column and row, and the square you chose exclude nine squares from being chosen next time.
There are 
remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.
Finally, the last square has 
remaining choices.
The number of ways to choose squares is 
but you must divide by 
since some sets are the same.
![\[\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } 600}\]](http://latex.artofproblemsolving.com/c/c/1/cc16a325891282648c5127af09cca9adef42b279.png)
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
