Suppose n different games are to be given to n children. In how many ways can this be done so that eactly one child gets no game.

Answer

To solve the problem of distributing n different games to n children such that exactly one child receives no game, we can follow these steps:

### Step-by-Step Solution:

1. **Choose the Child Who Will Not Receive a Game**:
We need to select one child out of the n children who will not receive any game. The number of ways to choose one child from n children is given by:
\[ \binom{n}{1} = n \]

2. **Distribute the Games to the Remaining Children**:
After choosing one child who will not receive a game, we have \(n-1\) children left. We need to distribute \(n\) games among these \(n-1\) children. Since one child must receive 2 games, we can choose one of the \(n-1\) children to receive 2 games. The number of ways to choose which child will receive 2 games is:
\[ \binom{n-1}{1} = n-1 \]

3. **Choose 2 Games for the Selected Child**:
Now, we need to select 2 games out of the n games to give to the chosen child. The number of ways to choose 2 games from n games is given by:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]

4. **Distribute the Remaining Games**:
After giving 2 games to one child, we have \(n-2\) games left to distribute among \(n-2\) children. Each of these children must receive exactly 1 game. The number of ways to distribute \(n-2\) games to \(n-2\) children is given by the number of permutations of \(n-2\) games, which is:
\[ (n-2)! \]

5. **Combine All Parts**:
Now, we can combine all the parts to find the total number of ways to distribute the games:
\[ \text{Total Ways} = n \times (n-1) \times \binom{n}{2} \times (n-2)! \]
Substituting the value of \(\binom{n}{2}\):
\[ \text{Total Ways} = n \times (n-1) \times \frac{n(n-1)}{2} \times (n-2)! \]

6. **Simplify the Expression**:
Simplifying the expression gives:
\[ = \frac{n \times (n-1)^2 \times n!}{2} \]

### Final Answer:
Thus, the total number of ways to distribute n different games to n children such that exactly one child gets no game is:
\[ \frac{n \times (n-1)^2 \times n!}{2} \]