# In how many ways can one divide 10 people into 4 unequally sized groups?

Many questions on this site involve counting the number of ways one can divide a set of n people into equally-sized groups, but how would one do so for unequally-sized groups?

The answers for this question don't provide an explanation of how to do this.

More specifically, what are the number of ways I can divide a set of 10 people in the following groups:

• Two groups will contain 3 people.
• One group will contain 2 people.
• One group will contain 1 person.

For this specific question, I was thinking

$$\frac{\dbinom{10}{3} \dbinom{7}{3} \dbinom{4}{2} \dbinom{2}{1}}{4!}$$

The numerator represents the grouping and the denominator represents the fact that the order of the grouping does not matter.

However, the answers to the referenced question show that this is incorrect.

• Since $3+3+2+1=9$, one of the $10$ people will not be in any group. Fine, but it's confusing to call that "dividing $10$ people into groups". – bof Oct 30 '13 at 4:10

Because the groups of $2$ and $1$ are distinct sizes, you don't have to consider reordering those groups. The denominator should be $2!=2$ because there are $2$ ways to order the two groups of three. The numerator is correct.
• No, the only other order that represents the same grouping is $\{\{DEF\},\{ABC\},\{GH\},\{I\}\}$ You can require that the doubleton and singleton come last-you can tell those sets apart by the number of elements. If you had $4+4+4+3+3+2+2$ the denominator would be $3!2!2!$ because you could reorder each size. – Ross Millikan Oct 30 '13 at 3:36
${(n-1)+r}\choose r$ for $r$ elements placed into one of $n$ groups. So if there were $10$ x's and $4$ groups it would look something like $x|xx|xx|xxxxx$ and be ${13\choose 10}=1716$. For this case you can never divide it into $4$ equally sized groups so do not need to worry about overcounting in that regard but if you did I believe it would just be the one case if $n$ evenly divides $r$. This would also count all cases where there are $0$ in a given group(s) so that would also be overcounting if you wanted at least $1$ per group.