SOLUTION TO PROBLEM H-5

There are

$$\frac{(40)!}{(2!)^{20}}$$

ways of dividing the 40 players into 20 ordered pairs of two each.(That is there are $\frac{(40)!}{2^{20}}$ ways of dividing the players into a first pair, a second pair, and so on.) Hence there are

$$\frac{(40)!}{(2!)^{20}(20)!}$$

ways of dividing the players into (unordered) pairs of 2 each. Furthermore, since a division will result in no offensive-defensive pairs if the offensive (and defensive) players are paired among themselves, it follows that there are

$$\left[\frac{(20)!}{(2!)^{10}(10)!}\right]^2$$

such divisions. Hence the probability of no offensive-defensive roommate pair is given by

$$P_0=\frac{\left[\frac{(20)!}{(2!)^{10}(10)!}\right]^2} {\frac{(40)!}{(2!)^{20}(20)!}}=\frac{[(20)!]^3}{[(10)!]^{2}(40)!}$$

To determine P_2i, the probability that there are 2ioffensive-defensive pairs, we first note that there are $\binom {20}{2i}^2$ways of selecting 2i offensive players and the 2i defensive players who are to be in the offensive-defensive pairs. These 4i players can then be paired up into (2i)! possible offensive-defensive pairs. As the remaining 20-2i offensives (and defensives) must be paired among themselves, it follows that there are

$$\binom {20}{2i}^2(2i)!\left[ \frac{(20-2i)!}{2^{10-i}(10-i)!}\right] ^2$$

divisions which lead to 2i offensive-defensive pairs. Hence

$$P_{2i} = \frac{\binom {20}{2i}^2(2i)!\left[ \frac{(20-2i)!}{2... ... ^2}{\frac{(40)!}{(2!)^{20}(20)!}} \qquad i = 0, 1, \dots , 10$$

The P_2i, $i = 0, 1, \dots , 10$ can now be computed or they can be approximated by making use of a result of Stirling which shows that n! can be approximated by $n^{n+{1\over 2}} + e^{-n}\sqrt{2\pi }$. For instance, we obtain that

$$P_0 \approx 1.3403 \times 10^{-6},$$

$$P_{10} \approx .345861,$$

$$P_{20} \approx 7.6068 \times 10^{-6}.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Thu Feb 14 20:34:55 CST 2002