In how many ways can we distribute $2$ types of gifts?
Solution 1:
Your reasoning for the problem where kids with no gifts are allowed is correct. To solve the main problem you need to apply the PIE to exclude cases where one or more kids get no gift. You will obtain:
$$ \sum_{i=0}^{k} (-1)^i\binom ki\binom{n+k-1-i}{n}\binom{m+k-1-i}{m}. $$