Problem Statement
$8$ children came to the court and played
- $4$ games of $\text{Game-1}$
- $3$ games of $\text{Game-2}$
- $10$ games of $\text{Game-3}$
Winner of each game gets gift. There are different gift to all games (Three types of gifts). How many options are there for getting gifts?
None of the children won more than $4$ games of $\text{Game-3}$.
My Attempt : I thought of taking all the possibilities and subtracting the possibility that someone would win $5,6,7,8,9,10$ but it seems too complicated.