here is my problem:
4 digits are picked randomly (from 0 to 9), in how many combinations can you have only 2 different digits? Meaning that I would have let us say 2,2,4,4 or 5,5,5,8.
So the way I wanted to do this was imagining that the digits were actually like 4 oranges going to 10 children and how many possibilities are there for only 2 children getting oranges. So I figured this would be separating the groups first so
10 choose 8 choose 1 choose 1 x 4 choose 2 choose 2
for the possibilities where both children get 2 oranges. And now for the possibilities where 1 of the children gets 3 and the other gets 1, would be
10 choose 8 choose 1 choose 1 x 4 choose 3 choose 1.
Unfortunately this gives me $$\frac{10!}{8!}\times \frac{4!}{2!2!}+\frac{10!}{8!}\frac{4!}{3!} = 900$$ possibilities whereas the right answer is $630.
Can anyone help me out please! and with this type of solving the problem preferably as I am trying to figure out the subtleties of the multinomial theorem!
As you point out, we can have a $2$-$2$ split or a $3$-$1$ split.
For $2$-$2$ splits, we choose $2$ numbers from $10$. There are $\binom{10}{2}=45$ ways to do this. For each choice, there are $\binom{4}{2}$ ways of arranging them in a row, total $270$.
For $3$-$1$ splits, the number we have $3$ of can be chosen in $10$ ways, and the number we have $1$ of can be chosen in $9$ ways. Then the digits can be arranged in $4$ ways, total $360$.
Remark: If we are not interested in counting ordered arrangements, then the total is $135$. To produce $630$, we need to take order into account.