Combinatorial - Ways to create subcommittees of a certain size out of a committee? Each member of a 10 member committee must be assigned to exactly one of 3 subcommittees (management, supervisor, employee). If these subcommittees are to contain 1,3, and 6 members respectively, how many different subcommittees can be appointed?
I solved this by doing this:
c(10,1) for management, c(9,3) for supervisor, c(6,6) for employees.
c(10,1) * c(9,3) * c(6,6) = different ways to arrange subcommittees. Is this correct? If not, what did I do wrong and what's a better way?
 A: Yes, indeed, your answer is correct. There is in fact a simpler way to approach this sort of problem: through the use of multinomial coefficients (see the two rightmost terms below): 
$$\binom{10}{1}\cdot\binom{9}{3} \cdot\binom 66 =\dfrac {10!}{\require{cancel}1!\cdot\color{blue}{\cancel{ 9!}}} \cdot \dfrac {\color{blue}{\cancel{9!}}}{\color{blue}{\cancel{6!}}\cdot 3!}\cdot \dfrac{\color{blue}{\cancel{6!}}}{6!} = \dfrac{10!}{1!\cdot 3!\cdot 6!} = \binom{10}{1, 3, 6}$$
Notice the nice cancellations in the factorials above! Do take a moment to read over the linked entry in Wikipedia: multinomial coefficients. Multinomial coefficients are perfectly suited to just this sort of problem, and simply save you some calculations.
A: Yup, that's right.  There's a few ways of doing it, all equivalent.
$\text{Y}{\small \text{OUR}}\text{ W}{\small \text{AY}}$: $${10 \choose 1}{9 \choose 3} = \frac{10!}{1!(10-1)!}\frac{9!}{3!(9-3)!} =840$$ There are $10$ choices for the manager.  After you pick him, there's only nine choices left for the three supervisors, and after that everybody else is an employee.
$\text{M}{\small \text{Y}}\text{ W}{\small \text{AY}}$:
$${10 \choose 6}{4 \choose 1}=\frac{10!}{6!(10-6)!}\frac{4!}{1!(4-1)!} =840$$
It's similar to the above: first pick the employees, then the manager.
Can you think of a third way? 
Moreover, can you generalize by finding a formula to count the ways to choose any number of subcommittees (of whichever sizes) from $n$ people?  (Hint: does anything cancel in the above formulas?)
