# Multisets with Exact Number of Repeated Integers

Given a multiset that contains 5 numbers where the numbers are from 0 to 5 inclusive, and the numbers can be repeated:

a) In how many ways can you have a multiset with exactly four 4s?

b) In how many ways can you have a multiset with exactly three 3s?

c) In how many ways can you have a multiset with exactly two 2s?

d) In how many ways can you have a multiset with exactly one 1s?

e) In how many ways can you have a multiset with exactly zero 0s?

I know that the number of repeated combinations of $k$ elements from $n$ available elements is given by the expression ${n+k-1 \choose k}$, but how would you apply it to this problem.

For example, for a) would it be ${5+3-1 \choose 3} = 35$

I am very confused on this subject, and require some help.

We do exactly two $2$'s, you can do the rest.
We need to count the number of ways to produce "the rest" of the multiset. So we need to count the number of $3$-element multisets, where the elements are chosen from the collection $\{0,1,3,4,5\}$.
You quoted a formula for counting the number of $3$-element multisets taken from a collection of $5$ numbers. It yields the answer $\binom{5+3-1}{3}$.
Remark: We certainly do not need the formula to deal with Question (a). For we need to choose one non-$4$ to go with the four $4$'s. There are $5$ numbers ot choose from, so we have $5$ choices.
The formula you mentioned does work, however, with $n=5$ and $k=1$.