Difference between revisions of "2005 Alabama ARML TST Problems/Problem 5"

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==Solution==
 
==Solution==
  
We choose two squares to be blue and one to be red; then the green's position is forced. There are <math>\binom{4}{2}\binom{2}{1}=12</math> ways to do this.
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We choose two squares to be blue and one to be red; then the green's position is forced. There are <math>{4 \choose 2}{2 \choose 1}=12</math> ways to do this.
  
 
*[[2005 Alabama ARML TST]]
 
*[[2005 Alabama ARML TST]]
 
*[[2005 Alabama ARML TST/Problem 4 | Previous Problem]]
 
*[[2005 Alabama ARML TST/Problem 4 | Previous Problem]]
 
*[[2005 Alabama ARML TST/Problem 6 | Next Problem]]
 
*[[2005 Alabama ARML TST/Problem 6 | Next Problem]]

Revision as of 14:48, 17 November 2006

Problem

A $\displaystyle 2\times 2$ square grid is constructed with four $\displaystyle 1\times 1$ squares. The square on the upper left is labeled A, the square on the upper right is labeled B, the square in the lower left is labled C, and the square on the lower right is labeled D. The four squares are to be painted such that 2 are blue, 1 is red, and 1 is green. In how many ways can this be done?

Solution

We choose two squares to be blue and one to be red; then the green's position is forced. There are ${4 \choose 2}{2 \choose 1}=12$ ways to do this.