Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer

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 $

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.