A counting problem?

The question is:

36 people are to be divided into groups of 6. Four of the boys insist on being in the same group as each other and 3 of the girls also want to be in the same group as each other. How many ways can the 36 people be divided into the groups?

This is how I went about answering it:

4 boys are put into one group. Two more individuals are required to make up a group of 6. 3 girls are put into another group. 3 more individuals are required to make up this group.

We have 29choose2 multiplied by 27choose3 to make up these two groups.

Then ofcourse the remaining 24 can be allocated to 4 indistinguishable groups by (24+6-1)choose6 i.e. 29choose6.

My final answer is 406+2925+475020 = 478,351

Just wanted some feedback as to whether I'm doing it right? Thanks.

Your answer is a gross, gross undercount.

There are ways to put twenty-four people into four groups of six each. These are known as unordered partitions.

Quote:

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Originally Posted by Plato

Your answer is a gross, gross undercount.

There are ways to put twenty-four people into four groups of six each. These are known as unordered partitions.

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Damn. My answer was completely off base then.

So what you've suggested is the same as (24C6 x 18C6 x 12C6 x 6C6)/4!. Thats how I went about it in the beginning.

Ok then so would then be multiplied by (29C2 x 27C3) equaling 1.14239514 x 10^17.

Infact I don't know why I added everything together in my OP. Stupid mistake. :(

That is now correct.

Thanks Plato :)