How many ways two types of items can be given? In how many ways, we can give $100$ type-$1$ items and $100$ type-$2$ items to $28$ people in such a way that each one receive at least one item and no one should get both type-$1$ and type-$2$ items.
My idea:
There are 28 people, so we can make the pairs $(1,27), (2,26) \cdots (27,1)$. That means, we are dividing $28$ people in such a way that sum of ordered pairs is $28$. Then the number of choices will be
$$ \sum_{k=1}^{27}\binom{100}{k} \binom{100}{28-k}.$$ Is it correct?
 A: You need $\binom{28}{k}$ to choose $k$ people to get type-1 items. The remaining $28-k$ people will get type-2 items.
Using the stars and bars formula $\binom{n-1}{k-1}$ for positive integers, the number of ways to distribute $100$ type-1 items to $k$ people so that each person gets at least one item is $\binom{99}{k-1}$, and the number of ways to distribute $100$ type-2 items to $28-k$ people so that each person gets at least one item is $\binom{99}{27-k}$.
All together, the total number of ways to distribute the items in the prescribed manner is $$\sum_{k=1}^{27}\binom{28}{k}\binom{99}{k-1}\binom{99}{27-k}$$
A: This is not an answer, but was too long to be a comment either. Hope it helps.
HINTS
HINT: How many possibilities does one person have for kind of item?
HINT 2: Since each object is of the same type, what can it be restated as if only that object is being distributed?

 Stars and Bars

To think about: How to deal with running out?
Eg: First person gets 28 of type 1, then how do deal with it?
