How many ways can 15 people be divided into 3 classes of 5, if there are 3 blond people....

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How many ways can 15 people be divided into 3 classes of 5, if there are 3 blond people and each class needs to have 1 blond person?

My attempt at the question:

First, the 3 blond people are excluded from the 15, leaving us with 12. To divide the 12 people into 3 groups of 5, where each group has 1 person already, we must do: 12C4 * 8C4 * 4C4

Now, this is where I get a little confused. I think that, because the groups are "distinct", due to the presence of 1 blond person in each group, there's no need to divide by 3!, despite the groups being of equal size.

Is this logic correct?

As a spin-off question, if all the blond people have to be in 1 class, then we'll have to do 12C2 (to choose the other 3 people for that class) * 10C5 * 5C5 divided by 2!, as the other two classes aren't distinct. Is this correct?

I don't have solutions to these questions so I cannot determine if it's right or wrong. Any advice will be much appreciated.

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Yes. The logic is correct.

For the first, each blonde person identifies their group.

For the second, only the group with them all is uniquely identified.