# How many ways $12$ persons may be divided into three groups of $4$ persons each?

How many ways $12$ persons may be divided into three groups of $4$ persons each?

I think the answer should be $\frac{12!}{(4!)^3}$ but the suggested correct answer is $5775$, could anybody explain where I am going wrong?

• Interestingly, this gives a proof that $\displaystyle \frac{(a_1 + a_2 + \dots + a_n)!}{a_1!a_2!\dots a_n!}$ is divisible by $\displaystyle n!$. – Aryabhata Feb 27 '12 at 18:48
• Ask yourself the following two questions: (1) How many ways are there to divide a group of $12$ people into $2$ volleyball teams of $6$ each, one team to wear blue, the other to wear red? (2) How many ways are there to divide $12$ people into two volleyball teams at a nudist camp? – André Nicolas Feb 27 '12 at 19:02
• @Aryabhata Except that $\displaystyle{\frac{(4+3)!}{4!3!}}$ (i.e., $n=2, a_1=4, a_2=3$) isn't divisible by $2!$ - you need interchangability of the pieces! I think the correct statement is that $\displaystyle{\frac{(na)!}{(a!)^n}}$ is always divisible by $n!$. – Steven Stadnicki Feb 27 '12 at 20:12
• @StevenStadnicki: Right! Thanks. – Aryabhata Feb 27 '12 at 20:16
• Any idea what the number would be in the case that 2 people aren't allowed to be together? – bibo_extreme Apr 3 '17 at 14:12

The answer is $\frac{12!}{(4!)^3\cdot3!}=5775$ because the $3!$ different orders of the three groups do not matter either, so your solution was almost correct.

We can also organize the count in a different way. First line up the people, say in alphabetical order, or in student number order, or by height.

The first person in the lineup chooses the $3$ people (from the remaining $11$) who will be on her team. Then the first person in the lineup who was not chosen chooses the $3$ people (from the remaining $7$) who will be on her team. The double-rejects make up the third team.

The first person to choose has $\binom{11}{3}$ choices. For every choice she makes, the second person to choose has $\binom{7}{3}$ choices, for a total of $$\binom{11}{3}\binom{7}{3}.$$

Remark: The lineup is a device to avoid multiple-counting the divisions into teams. The alternate (and structurally nicer) strategy is to do deliberate multiple counting, and take care of that at the end by a suitable division.

Since you're talking about people, order doesn't matter, so you have to add a $3!$ dividing there.

From $12$ persons , $4$ persons can be chosen in $\binom{12}{4}$ ways. From rest $(12-4)=8$ persons 4 persons can be chosen in $\binom{8}{4}$ ways.Remaining $4$ will form the third group.Thus $\binom{12}{4} \binom{8}{4}$ are the ways for form three groups of $4$ persons out of $12$ persons.But we have introduced order in group formation. Now three groups can be permuted $3!$ ways and they are all same.Hence correct number of ways $= \frac{\binom{12}{4} \binom{8}{4}}{3!}=5775$