I'm reading my Probability + Stats book for engineers and came across this example in the actual reading. The solution is given in the reading...but I don't understand why we have to divide in the numerator by 3! and the denominator by 6!.
A basketball team consists of 6 black and 6 white players. The players are to be paired in groups of two for the purpose of determining roommates. If the pairings are done at random, what is the probability that none of the black players will have a white roommate?
Let us start by imagining that the 6 pairs are numbered — that is, there is a first pair, a second pair, and so on. Since there are 12C2 different choices of a first pair; and for each choice of a first pair there are 10C2 different choices of a second pair; and for each 2 choice of the first 2 pairs there are 8C2 choices for a third pair; and so on, it follows from the generalized basic principle of counting that there are (12c2)* (10c2)* (8c2)* (6c2)* (4c2)* (2c2) = 12!/(2!)^6 ways of dividing the players into a first pair, a second pair, and so on. Hence
there are (12)!/((2^6)6!) ways of dividing the players into 6 (unordered) pairs of 2 each. Furthermore, since there are, by the same reasoning, 6!/((2^3)3!) ways of pairing the white players among themselves and 6!/((2^3)3!) ways of pairing the black players among themselves, it follows that there are (6!/((2^3)3!)^2 pairings that do not result in any black–white roommate pairs.
Hence, if the pairings are done at random (so that all outcomes are equally likely), then the desired probability is
note: % means divide / also means divide
6!/(*(2^3)3!)% ((12)!/((2^6)6!))= 5/231 =.0216*
Hence, there are roughly only two chances in a hundred that a random pairing will not result in any of the white and black players rooming together.