SOLUTION: A bag contains n white and n red balls.pairs of balls are drawn without replacement until the bag is empty.If the number of ways in which each pair consists of one red and one whit

Question 1136293: A bag contains n white and n red balls.pairs of balls are drawn without replacement until the bag is empty.If the number of ways in which each pair consists of one red and one white ball is 14,400,then what is the value of n?
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

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.

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.

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.

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

(n^2)*(n-1)^2*...*(2)^2*(1)^2 = (n!)^2

In this problem, that number of ways is 14,400:

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