`n` different toys have to be distributed among `n` children. Find the number of ways in which these toys can be distributed so that exactly one child gets no toy.

Options

A. `n!`
B. `n!.^(n)C_(2)`
C. `(n-1)!.^(n)C_(2)`
D. `n!.^(n-1)C_(2)`

Answer

To solve the problem of distributing `n` different toys among `n` children such that exactly one child receives no toy, we can follow these steps:

### Step-by-Step Solution:

1. **Select the Child Who Gets No Toy**:
We need to choose one child out of `n` children who will not receive any toys. The number of ways to choose 1 child from `n` is given by:
\[ \text{Ways to choose the child} = \binom{n}{1} = n \]

2. **Distribute Toys Among Remaining Children**:
After selecting the child who will not receive a toy, we have `n - 1` children left. We need to distribute `n` toys among these `n - 1` children.

3. **Choose One Child to Receive Two Toys**:
Since we have `n` toys and `n - 1` children, one child must receive 2 toys while the others receive 1 toy each. We can choose which of the `n - 1` children will receive 2 toys. The number of ways to choose 1 child from `n - 1` is:
\[ \text{Ways to choose the child for 2 toys} = \binom{n-1}{1} = n - 1 \]

4. **Select 2 Toys for the Chosen Child**:
Now, we need to select 2 toys from the `n` available toys to give to the chosen child. The number of ways to choose 2 toys from `n` is:
\[ \text{Ways to choose 2 toys} = \binom{n}{2} = \frac{n(n-1)}{2} \]

5. **Distribute Remaining Toys**:
After giving 2 toys to one child, we have `n - 2` toys left to distribute among the remaining `n - 2` children. Each of these children will receive exactly 1 toy. The number of ways to distribute `n - 2` toys to `n - 2` children (where each child gets exactly one toy) is given by:
\[ \text{Ways to distribute remaining toys} = (n - 2)! \]

6. **Combine All the Steps**:
Now, we can combine all the ways calculated in the previous steps:
\[ \text{Total ways} = \binom{n}{1} \times \binom{n-1}{1} \times \binom{n}{2} \times (n - 2)! \]
Substituting the values we calculated:
\[ \text{Total ways} = n \times (n - 1) \times \frac{n(n - 1)}{2} \times (n - 2)! \]

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

Since \((n - 2)! = \frac{n!}{n(n-1)(n-2)}\), we can rewrite the total ways as:
\[ = \frac{n!}{2} \]

### Final Answer:
The total number of ways to distribute `n` different toys among `n` children such that exactly one child gets no toy is:
\[ \frac{n!}{2} \]