How many ways are there first to pick a subset of n numbers from 50 distinct numbers, and next pick a second subset of k numbers such that each number in the first subset is less than each number in the second subset
I will use some new letters to discuss general counting results, to avoid confusion.
The number of ways to select v elements from a set of u elements is simply $\displaystyle \binom{u}{v}$, evidenced by the fact that it's acceptable to say out loud "u choose v". What you wrote is the number of multisets of cardinality v whose elements belong to a set with u (distinct) elements, $\displaystyle \binom{u+v-1}{v}$. From the wording of the problem, we need $\displaystyle \binom{u}{v}$.
So, obviously if n+k>50 then the answer is 0. Assume n+k is less than or equal to 50. We can also assume n and k are positive, and if need be consider the cases n=0 or k=0 as special (trivial) cases.
Assume for simplicity that the set with 50 elements is {1,...,50}.
Then the least element of the k-subset can range from n+1 to 50-k+1.
Fix the least element of the k-subset and name it i. For the k-subset, we have k-1 elements remaining, to be chosen from 50-i elements. For the n-subset, we have n elements to be chosen from i-1 elements.
Then take the sum as i goes from n+1 to 50-k+1.
(Edit: Sorry, originally I had n and k switched throughout, due to misreading the problem.)