# Probability/Combinatorics Question

## Homework Statement

The NHL currently has a total of 30 teams in 4 divisions (7 teams in the Pacific
Division, 7 in the Central Division, 8 in the Metropolitan Division, and 8 in the
Atlantic Division). Suppose the NHL gets a new commissioner, and they have
the curious notion of reshuffling teams by randomly assigning the 30 teams to the
divisions (leaving the number of teams in each division the same as above). How
many different ways can this be done?

None.

## The Attempt at a Solution

There are 30 teams and they must be partitioned into teams of 7, 7, 8, 8, which represent the four divisions..

$$(30!)/(7!7!8!8!)$$ using a partitioning rule..

I'm not sure if this is the right way to go about it, any hints would be appreciated.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

The NHL currently has a total of 30 teams in 4 divisions (7 teams in the Pacific
Division, 7 in the Central Division, 8 in the Metropolitan Division, and 8 in the
Atlantic Division). Suppose the NHL gets a new commissioner, and they have
the curious notion of reshuffling teams by randomly assigning the 30 teams to the
divisions (leaving the number of teams in each division the same as above). How
many different ways can this be done?

None.

## The Attempt at a Solution

There are 30 teams and they must be partitioned into teams of 7, 7, 8, 8, which represent the four divisions..

$$(30!)/(7!7!8!8!)$$ using a partitioning rule..

I'm not sure if this is the right way to go about it, any hints would be appreciated.

Sure, and to convince yourself you can do it sequentially. Call the divisions 1--4. In how many distinct ways can you assign teams to division 1? (That is, we pick the 7 to go into division 1 and the remaining 23 go into non-1.) For each distinct division-1 assignment, in how many different ways can we assign 7 to division 2? (That is, of the 23 still left, we assign 7 to division 2 and the other 16 to not [1 or 2].) The first division can be picked in
$$N_1 = \binom{30}{7} = \frac{30!}{7! \; 23!}$$
different ways. For each such assignment the second division can be picked in
$$N_2 = \binom{23}{7} = \frac{23!}{7! \; 16!}$$
different ways. Together, divisions 1 and 2 can be assigned in
$$N_{12} = N_1 \: N_2 = \frac{30!}{7! \; 23!} \cdot \frac{23!}{7! \; 16!} = \frac{30!}{ 7! \; 7! \; 16!}$$
different ways. Keep going like that until all divisions have been assigned, and you will get your suggested solution.

1 person
Thanks for the reassurance, I appreciate it. I don't have much experience in the math stats setting as I recently switched into the Statistics major, but I think I'm getting the hang of it.