# Thread: Counting Principles

1. ## Counting Principles

A homework question:

Twelve children are divided up into three groups, of five, four, and three. In how many ways can this be done if the order within each group is not important?

I'm not quite sure how to attack this question.

Any suggestions?

Thanks!

2. Originally Posted by strgrl
A homework question:

Twelve children are divided up into three groups, of five, four, and three. In how many ways can this be done if the order within each group is not important?

I'm not quite sure how to attack this question.

Any suggestions?

Thanks!
${12 \choose 5} {(12-5) \choose 4} {(12 - 5 - 4) \choose 3} = {12 \choose 5} {7 \choose 4} {3 \choose 3}$ ......

3. Hello, strgrl!

Twelve children are divided up into three groups, of five, four, and three.
In how many ways can this be done if the order within each group is not important?
This is a "partition" problem.
Evidently, you haven't been introduced to them yet.

The answer is: . ${12\choose5,4,3} \;=\;\frac{12!}{5!4!3!} \;=\;27,720$

4. Thank you!