Eleven students into two groups of four and one of three The problem: 11 students are to be divided into 3 groups, 2 groups of 4 students and 1 group of 3 students. In how many ways can I do this?
The order of the groups and the order of the students in every group doesn't matter
I just followed the formula and got
$ 11!\choose  4! (11-4)!$*
$ 7!\choose  4! (7-4)!$*
$3!\choose 3!(3-3)!$
and thought i solved it. So our teacher did it in class afterwards and he decided it should be
$11\choose 4$ *
 $7\choose 4$ *
 $3\choose 3$ /2 
my question is why is it /2 at the end? is it because there are 2 groups that are equal?
Could someone show me the correct way to do it? So I got it clarified.
Thanks
 A: It should have been mentioned in the question that the groups are unlabelled, i.e. indistinguishable, except by size.
The reason may be clearer if we consider one such division
We would consider $ABCD\;|\; EFGH | IJK $, and, say $EFGH\; |IJK\;|ABCD $ to be identical divisions.
Using permutations, we would write the answer as $\;\dfrac{11!}{(4!4!3!)\times(2!)}$
The denominator's first part removes permutations within groups,
and the $2!$ removes those between indistinguishable groups
It could, of course, instead be written as the book has done, viz. $\binom{11}4\binom7{4}\binom33 /2$ 
A: Here is another way to get the answer:  Line the $11$ students up from left to right.  Pick $3$ of them for the group of three, which can be done in $11\choose3$ ways.  Of the remaining students, the leftmost one must go into a group with $3$ of the other $7$, who can be chosen in $7\choose3$ ways.  So the total number of ways the groups can be formed is
$${11\choose3}{7\choose3}$$
A: The answer calls for the use of multinomial coefficients. See e.g. http://mathworld.wolfram.com/MultinomialCoefficient.html for more.
If we have a set of $n$ elements to be partitioned into $k$ subsets with $n_1, \ldots, n_k$ elements, respectively (where $\sum^m_{i=1} n_i = n$), then there are
$$ \binom{n}{n_1, \ldots, n_k} = \frac{n!}{n_1! \cdots n_k!} $$
ways to do this. The number $\binom{n}{n_1, \ldots, n_k}$ is called a multinomial coefficient.
