Question

math

Answer 1

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

Answer 2

did that help?

Answer 3

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

Answer 4

9 choose 3

Answer 5

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

Answer 6

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

Answer 7

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
\[\large \frac{9!}{3!3!3!}\]

Answer 8

Thank you so much guys.. =)

Answer 9

You're welcome :)