# Thread: one more committee problem

1. ## one more committee problem

We are going to form 3 committees from the people in our class, 4 people per committee. There are 12 people in our class and everyone will be placed on exactly one committee. 4 ppl on com. X, 4 on Y and 4 on Z. How many options for committee assignments are there?

i think this is 12C4 * 8C4 * 4C4?

2. Originally Posted by ihavvaquestion
We are going to form 3 committees from the people in our class, 4 people per committee. There are 12 people in our class and everyone will be placed on exactly one committee. 4 ppl on com. X, 4 on Y and 4 on Z. How many options for committee assignments are there?

i think this is 12C4 * 8C4 * 4C4?
Looks spot on to me. If you only had 8 people to start with and had to form two 4 people committees there would be 8 ways to choose the first member of the first committee, then 7 ways to choose the second, then 6 to choose the third, then 5 to choose the last member of the first committee which corresponds to 8C4 and of course 4C4 is just 1. Now just add the 12C4 apply the product principle and get your answer!

3. can anyone confirm this to be the correct answer?

4. Please forgive me for misunderstanding the question.

5. The answer is indeed $\binom{12}{4}\binom{8}{4}\binom{4}{4}=\frac{12!}{( 4!)^3}$.
These are ordered partitions because the committees have names.
Also note that this is the number of ways to rearrange the string $XXXXYYYYZZZZ$.