This is a very basic counting problem, however I couldn't recall my memory to understand the answer correctly.
From "A First Course in Probability by Sheldon Ross",
Example A football team consists of 20 offensive and 20 defensive players. The players are paired in groups of 2 for the purpose of determining roommates....
There are $\dfrac{40!}{(2!)^{20}}$ ways of dividing 40 players into 20 "ordered pairs" of two each.
I tried a smaller set, says:
- Offensive = ${a_1, a_2}$
- Defensive = ${b_1, b_2}$
so all ordered pairs are : $a_1a_2, a_2a_1, b_1b_2, b_2b_1$ $a_1b_1, b_1a_1, a_1b_2, b_2a_1$ $a_2b_1, b_1a_2, a_2b_2, b_2a_2$ So there are 12 of pairs. Now if I use the formula was given above, I got: $\dfrac{4!}{(2!)^2} = \dfrac{4.3.2}{4} = 6$ which is clearly incorrect.
So my question is, should the formula given from the Example be $\dfrac{40!}{{2!}^{19}}$?