# Thread: How many ways could a class of 18 students divide into group

1. ## How many ways could a class of 18 students divide into group

How many ways could a class of 18 students divide into groups of 3 students each?

Have 18 students take 3 out, then do that 6 times?
6(18P3)

is that right?

Thanks

2. Try $\L\frac{{18!}}{{\left( {6!} \right)\left( {3!} \right)^6 }}$

Can you explain why that works?

EDIT:
I first read it as three groups not three in a group.

3. ## Re: How many ways could a class of 18 students divide into g

Hello, wind!

How many ways could a class of 18 students divide into groups of 3 students each?

I'm sure that pka meant: $\L\:\frac{18!}{6!(3!)^6}$

Let's baby-step through this problem . . .

First, we'll number the groups, 1 to 6.

For group #1, there are: $\,{18 \choose 3}$ ways.
For group #2, there are: $\,{15 \choose 3}$ ways.
For group #3, there are: $\,{12 \choose 3}$ ways.
For group #4, there are: $\,{9 \choose 3}$ ways.
For group #5, there are: $\,{6 \choose 3}$ ways.
For group #6, there are: $\,{3 \choose 3}$ ways.

Hence, there are: $\L\:\frac{18!}{3!\sout{15!}}\cdot\frac{\sout{15!}} {3!\sout{12!}}\cdot\frac{\sout{12!}}{3!\sout{9!}}\ cdot\frac{\sout{9!}}{3!\sout{6!}}\cdot\frac{\sout{ 6!}}{3!\sout{3!}}\cdot\frac{\sout{3!}}{3!\sout{0!} } \:=\:\frac{18!}{(3!)^6}$ ways.

But the groups are not numbered; their order is not considered.

Since our answer includes the $6!$ different orders of the six groups,
. . we must divide by $6!$

Answer: $\L\:\frac{18!}{6!(3!)^6}$

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