Question 1136293
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There are initially n balls of each color.  We have to choose 1 of the n white balls and 1 of the n red balls.  The number of ways of doing that is "n choose 1" times "n choose 1", which is n*n = (n)^2.<br>
After that, there are (n-1) balls of each color.  We have to choose 1 of the n-1 white balls and 1 of the n-1 red balls.  The number of ways of doing that is "n-1 choose 1" times "n-1 choose 1", which is (n-1)(n-1) = (n-1)^2.<br>
We continue that until there just two balls left, one of each color; there is only 1 way to choose a pair from those two balls.<br>
The total number of ways of choosing the n balls of each color, each time taking a pair consisting of one white and one red, is then<br>
(n^2)*(n-1)^2*...*(2)^2*(1)^2 = (n!)^2<br>
In this problem, that number of ways is 14,400:<br>
{{{(n!)^2 = 14400}}}
{{{n! = 120}}}
{{{n = 5}}}