# Counting Principles

• April 2nd 2008, 02:46 PM
strgrl
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!
• April 2nd 2008, 04:06 PM
mr fantastic
Quote:

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!

${12 \choose 5} {(12-5) \choose 4} {(12 - 5 - 4) \choose 3} = {12 \choose 5} {7 \choose 4} {3 \choose 3}$ ......
• April 2nd 2008, 06:49 PM
Soroban
Hello, strgrl!

Quote:

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

• April 2nd 2008, 08:23 PM
strgrl
Thank you!