Please check my solution of distributing $4$ prizes among $6$ people 
There are $6$ people and $4$ identical prizes. Assuming that $0$ distribution of prizes is allowed, find the number of ways of distributing the $4$ identical prizes to the $6$ persons such that a person can get any number of prizes.

My answer is coming $246$ or $$\binom{6}{1}+\binom{6}{2}\cdot3+\binom{6}{3}\cdot3+\binom{6}{4}$$ I first picked up a person and found the number of ways of distributing all prizes to him. Then I picked up any two persons and found the number of ways of distributing all prizes to them. I continued this till the $4$th person as after that the prizes end. Is my method and my answer right$?$
 A: Your method is actually correct! This approach is great because the cases of $1$ person receiving all the presents, then $2$, $3$, and $4$ people are all separate (disjoint). With some care we can see why your method is correct.
The case with one person receiving all the presents can be represented as the ordered set $(4, 0)$.
The case with $2$ people can be represented as $(3, 1), (2, 2), (1, 3)$. $(0, 4)$ cannot happen as that is included in the case where any one person receives all the presents.
The case with $3$ people can be represented as $(2, 1, 1), (1, 2, 1), (1, 1, 2)$. As you might have noticed by now, none of these numbers can be $0$ or else it would be the same as one of the earlier cases. Therefore we must have $(1, 1, 1)$ already, and we have $3$ choices to add an extra $1$ to each entry.
I find this method more efficient however: realise that this is just a stars and bars problem. We have $4$ prizes as before, but also $5$ dividers to separate $6$ people, which makes $9$ objects in total. For instance, this is one partition of the $4$ prizes for $6$ people:
$$P| \ | PP | \ | \ | P$$
Since all of these objects are indistinguishable, we must use combinations and so this method gives:
$${9 \choose 4} = \boxed{126}.$$
