Number of ways in which 9 different toys be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more, is

Options

A. `((5!)^(2))/(8)`
B. `(9!)/(2)`
C. `(9!)/(3!(2!)^(3))`
D. none

Answer

To solve the problem of distributing 9 different toys among 4 children, where the distribution among the 3 elder children is even and the youngest child receives one toy more, we can follow these steps:

### Step 1: Determine the distribution of toys
We have 4 children (let's call them C1, C2, C3, and C4). The 3 elder children (C1, C2, and C3) will receive the same number of toys, and the youngest child (C4) will receive one toy more.

Let the number of toys received by each of the elder children be \( x \). Then, the youngest child will receive \( x + 1 \) toys.

Since there are a total of 9 toys, we can set up the equation:
\[ 3x + (x + 1) = 9 \]
This simplifies to:
\[ 4x + 1 = 9 \]
\[ 4x = 8 \]
\[ x = 2 \]

Thus, the distribution of toys will be:
- C1: 2 toys
- C2: 2 toys
- C3: 2 toys
- C4: 3 toys

### Step 2: Calculate the number of ways to distribute the toys
Now, we need to calculate the number of ways to distribute these toys. We will do this in stages:

1. **Select 2 toys for C1 from 9 toys**:
The number of ways to choose 2 toys from 9 is given by the combination:
\[ \binom{9}{2} \]

2. **Select 2 toys for C2 from the remaining 7 toys**:
After giving toys to C1, we have 7 toys left. The number of ways to choose 2 toys for C2 is:
\[ \binom{7}{2} \]

3. **Select 2 toys for C3 from the remaining 5 toys**:
After giving toys to C2, we have 5 toys left. The number of ways to choose 2 toys for C3 is:
\[ \binom{5}{2} \]

4. **Assign the remaining 3 toys to C4**:
C4 will receive the remaining 3 toys, and there is only one way to do this since all remaining toys go to C4.

### Step 3: Calculate the total number of distributions
The total number of ways to distribute the toys can be calculated by multiplying the combinations:
\[ \text{Total Ways} = \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2} \times 1 \]

Calculating each combination:
\[ \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 \]
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

Thus, the total number of ways is:
\[ \text{Total Ways} = 36 \times 21 \times 10 \]

### Step 4: Final Calculation
Calculating the product:
\[ 36 \times 21 = 756 \]
\[ 756 \times 10 = 7560 \]

### Final Answer
The total number of ways to distribute the 9 different toys among the 4 children is **7560**.

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