2822596

A basketball team consists of 6 frontcourt and 4 backcourt players

https://math.stackexchange.com/questions/2822596/understanding-solution-to-probability-problem/

A basketball team consists of 6 frontcourt and 4 backcourt players. If players are divided into roommates at random, what is the probability that there will be exactly two roommate pairs made up of a backcourt and a frontcourt player?

$$ Room = \mathit 2[X+Y] = \mathit 2[X] + tX \cdot Y + \mathit 2[Y] $$

If the rooms are labelled :

$$ Team = Room^5 $$

$$Team \rightarrow \left( { x^2 \over 2 } + txy + { y^2 \over 2 } \right)^5$$

$$ coeff( { x^6 \over 6! } ,{ y^4 \over 4! }) = 43200t^4+64800t^2+5400 $$

If the rooms are not labelled:

$$ Team = \mathit 5[Room] $$

$$ Team \rightarrow {1 \over 5!} \left( { x^2 \over 2 } + txy + { y^2 \over 2 } \right)^5$$

$$ coeff( { x^6 \over 6! } ,{ y^4 \over 4! }) = 360t^4+540t^2+45 $$

the types equation is $$ 360+540+45 = 945$$