## anonymous one year ago math

1. Oleg3321

well isnt it three students just go into each of the three groups so then all the nine students are split up evenly?

2. Oleg3321

did that help?

3. anonymous

I do not understand.. Hmm is it 9c3?

4. zzr0ck3r

9 choose 3

5. zzr0ck3r

$$\frac{9!}{3!(9-3)!}$$

6. zzr0ck3r

$$\dfrac{n!}{k!(n-k)!}$$ tells you how many ways you can choose k things from n things.

7. kropot72

The number of ways 9 students can be partitioned into three teams containing 3 students each is found as follows: There are 9 choices for the first student of team A, 8 choices for the second student of team A and 7 choices for the third student of team A. The order of the choices does not matter. Therefore the number of ways of choosing team A is given by $\large \frac{9\times8\times7}{3!}$ Having chosen team A, the number of ways of choosing team B is given by $\large \frac{6\times5\times4}{3!}$ and the number of ways of choosing team C is given by $\large \frac{3\times2\times1}{3!}$ Finally the total number of ways is given by $\large \frac{9!}{3!3!3!}$

8. anonymous

Thank you so much guys.. =)

9. kropot72

You're welcome :)