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.


The divisors arise because of the need to convert the number of ordered pairs into the number of unordered pairs. The difference is perhaps best illustrated by means of an example.

Consider having just 4 players: $A, B, C, D$ which are to be arranged into 2 pairs.

Using the logic of the quoted "solutions" there are $4C2 \times 2C2 = 6 \times 1 = 6$ ordered pairings. These 6 possibilities are

  1. $(AB,CD)$
  2. $(AC,BD)$
  3. $(AD,BC)$
  4. $(BC, AD)$
  5. $(BD, AC)$
  6. $(CD, AB)$

where $(x, y)$ means $x$ is the first pair and $y$ is the second pair.

However, if we ignore the order in which the pairs occur there are in reality only 3 possibilities since 1. and 6. both split the four players into identical pairings, as do 2. and 5., as do 3. and 4. Thus for 2 pairs of players there are 6 different pairings when pairings are ordered, but only 3 different pairings when the order in which the pairs occur is ignored.

More generally, when there are $n$ pairs of players, any specific set of $n$ pairs can be ordered in $n!$ ways ($n$ choices of which pair is first, $n-1$ for which pair is second, etc). Since this is true for any set of $n$ pairs it follows that for $n \times 2$ players the number of possible ordered pairs is equal to number of possible unordered pairs multiplied by $n!$.

The population on which the probability calculation (numerator divided by denoinator) is based is that of unordered pairs of players. However, the calculation of $12C2 \times 10C2 \times 8C2 \times 6C2 \times 4C2 \times 2C2$ (and the unstated calculation for black-black and white-white pairings of $6C2 \times 4C2 \times 2C2$) calculates the number of ordered pairs.

Since, the 12 players are divided into 6 pairs, the number of possible unordered pairs (denominator of the probability calculation) is calculated by dividing the number of ordered pairs by $6!$. Similarly, the numerator is concerned with number of possible white-white and black-black pairings so involves 6 players in 3 pairs and therefore $3!$ is used as a divisor.

Incidentally, the solution given should have squared the numerator so that the probability is $$\frac{(\frac{6!}{2^3 3!})^2}{\frac{12!}{2^6 6!}}$$ However, the numerical value of 5/231 is consistent with the corrected formula.


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