Let $S$ be a multi-set with k distinct different elements with given repetition numbers $n_1,n_2,\ldots,n_k$ respectively. Let $r$ be a positive integer such that there exists at least one $r$-combination of the set $S$. Show that in applying the inclusion-exclusion principle to determine the number of $r$-combinations of the set $S$, one has $A_1\cap A_2\cap\ldots\cap A_k=\varnothing$.
What I've done so far is noticed that since there exists one integer $r$ such that there is at least one $r$-combination of the set $S$, then $\sum_{i=1}^k n_i\ge k$. Also notice that each $n_i\ge1$. I can see that if one of those $n_i$ is $2$ or more, then I can see that the proposition is true. However what I'm having trouble with is the case when $n_1+n_2+\ldots+n_k=r$, because in this case $A_1\cap A_2\ldots\cap A_k=1$, correct? I'm unsure of how to proceed with this this proof any help is appreciated!
From "there is at least one r-combination of S", we can see that r is smaller or equal to n1 + n2 + ... + nk. When trying to figure out the cases of all the intersections, we need to assume there're n1+1, n2+1, ..., nk+1 elements respectively, which is clearly bigger than r. Hence, the intersection is the empty set.