# A counting problem?

• Nov 28th 2010, 11:02 AM
Markhor
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.
• Nov 28th 2010, 11:19 AM
Plato
Your answer is a gross, gross undercount.

There are $\dfrac{24!}{(6!)^4(4!)}$ ways to put twenty-four people into four groups of six each. These are known as unordered partitions.
• Nov 28th 2010, 11:38 AM
Markhor
Quote:

Originally Posted by Plato
Your answer is a gross, gross undercount.

There are $\dfrac{24!}{(6!)^4(4!)}$ ways to put twenty-four people into four groups of six each. These are known as unordered partitions.

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 $\dfrac{24!}{(6!)^4(4!)}$ 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. :(
• Nov 28th 2010, 12:05 PM
Plato
That is now correct.
• Nov 28th 2010, 12:13 PM
Markhor
Thanks Plato :)