# Math Help - ordered and unordered partitions

1. ## ordered and unordered partitions

In how many ways can 10 students be divided into three teams, one containing 4 students and the others 3?

Method 1 makes sense, in which the answer is $\frac{10!}{4!3!3!}\cdot\frac{1}{2}$.

But method 2 bothers me, where the answer is $\binom{10}{4}\binom{5}{2}$.

I can understand choosing 4 out of 10 the first time, but what is the logical explanation for choosing 2 out 5 the second time? This does not seem very consistent.

2. Originally Posted by novice
In how many ways can 10 students be divided into three teams, one containing 4 students and the others 3?

Method 1 makes sense, in which the answer is $\frac{10!}{4!3!3!}\cdot\frac{1}{2}$.

But method 2 bothers me, where the answer is $\binom{10}{4}\binom{5}{2}$.

I can understand choosing 4 out of 10 the first time, but what is the logical explanation for choosing 2 out 5 the second time? This does not seem very consistent.
Imagine you're the coach for this squad. You pick four of the ten students for the first team ( $\textstyle{10\choose4}$ ways of doing that). You then designate one of the other six students (doesn't matter who it is) to be a team captain, with instructions to choose two of the remaining five students to complete his/her team ( $\textstyle{5\choose2}$ ways of doing that). The three left-over students form the third team, of course.

3. Originally Posted by Opalg
Imagine you're the coach for this squad. You pick four of the ten students for the first team ( $\textstyle{10\choose4}$ ways of doing that). You then designate one of the other six students (doesn't matter who it is) to be a team captain, with instructions to choose two of the remaining five students to complete his/her team ( $\textstyle{5\choose2}$ ways of doing that). The three left-over students form the third team, of course.
Thank you, Opalg, for taking time answering my question. Your explanation makes a lot of sense. We need more of your kind to write math books.