10 different toys are to be distributed among 10 children. Total number of ways of distributing these toys so that exactly 2 children do not get any toy, is equal to:

Options

A. `(10!)^(2)((1)/(3!2!7!)+(1)/((2!)^(5)6!))`
B. `(10!)^(2)((1)/(3!2!7!)+(1)/((2!)^(4)6!))`
C. `(10!)^(2)((1)/(3!7!)+(1)/((2!)^(5)6!))`
D. `(10!)^(2)((1)/(3!7!)+(1)/((2!)^(4)6!))`

Answer

To solve the problem of distributing 10 different toys among 10 children such that exactly 2 children do not receive any toys, we can break down the solution into steps.

### Step-by-Step Solution:

1. **Identify the Total Number of Children and Toys**:
We have 10 different toys and 10 children. We need to ensure that exactly 2 children do not receive any toys.

2. **Choose the Children Who Will Not Receive Toys**:
We need to select 2 out of the 10 children to not receive any toys. The number of ways to choose 2 children from 10 is given by the combination formula:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \cdot 8!} = 45 \]

**Hint**: Use the combination formula to choose the children who will not receive toys.

3. **Distribute Toys to the Remaining Children**:
After selecting 2 children who will not receive toys, we have 8 children left to distribute the 10 toys. Since all toys are different, each of the 10 toys can go to any of the 8 children. The total number of ways to distribute 10 different toys to 8 children is given by:
\[ 8^{10} \]
This is because each toy has 8 choices (the remaining children).

**Hint**: Remember that each toy can go to any of the remaining children, leading to an exponential distribution.

4. **Calculate the Total Number of Ways**:
Now, we combine the two parts of our solution:
- The number of ways to choose the 2 children who do not receive toys: \( \binom{10}{2} = 45 \)
- The number of ways to distribute the toys among the remaining 8 children: \( 8^{10} \)

Therefore, the total number of ways to distribute the toys is:
\[ \text{Total Ways} = \binom{10}{2} \times 8^{10} = 45 \times 8^{10} \]

**Hint**: Multiply the number of ways to choose the children with the number of ways to distribute the toys.

5. **Final Calculation**:
To find the numerical value of \( 8^{10} \):
\[ 8^{10} = (2^3)^{10} = 2^{30} \]
Thus, the total number of ways becomes:
\[ 45 \times 2^{30} \]

**Hint**: Simplify \( 8^{10} \) using powers of 2 for easier calculations.

### Final Answer:
The total number of ways to distribute the toys such that exactly 2 children do not get any toy is:
\[ 45 \times 8^{10} \]