The number of ways in which 8 distinguishable apples can be distributed among 3 boys such that every boy should get atleast 1 apple & atmost 4 apples is `K P(7,3)`, where `K` has the value equal to

Options

A. 88
B. 66
C. 44
D. 22

Answer

To solve the problem of distributing 8 distinguishable apples among 3 boys such that each boy receives at least 1 apple and at most 4 apples, we can break down the solution into several steps.

### Step-by-Step Solution:

1. **Define the Variables**:
Let \( a_1, a_2, a_3 \) be the number of apples received by Boy 1, Boy 2, and Boy 3 respectively. We need to satisfy the conditions:
\[ a_1 + a_2 + a_3 = 8 \]
with the constraints:
\[ 1 \leq a_i \leq 4 \quad \text{for } i = 1, 2, 3 \]

2. **Transform the Variables**:
To simplify the problem, we can define new variables:
\[ b_1 = a_1 - 1, \quad b_2 = a_2 - 1, \quad b_3 = a_3 - 1 \]
This transformation ensures that \( b_i \geq 0 \). The equation now becomes:
\[ (b_1 + 1) + (b_2 + 1) + (b_3 + 1) = 8 \implies b_1 + b_2 + b_3 = 5 \]
with the new constraints:
\[ 0 \leq b_i \leq 3 \]

3. **Use the Stars and Bars Method**:
Without the upper limit, the number of non-negative integer solutions to \( b_1 + b_2 + b_3 = 5 \) can be calculated using the stars and bars method:
\[ \text{Number of solutions} = \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} = 21 \]

4. **Subtract Invalid Cases**:
Now we need to subtract the cases where any \( b_i > 3 \). Assume \( b_1 > 3 \). Let \( b_1' = b_1 - 4 \) (so \( b_1' \geq 0 \)). The equation becomes:
\[ b_1' + b_2 + b_3 = 1 \]
The number of non-negative integer solutions is:
\[ \binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3 \]
Since the same reasoning applies to \( b_2 \) and \( b_3 \), we subtract 3 for each case:
\[ \text{Total invalid cases} = 3 \times 3 = 9 \]

5. **Calculate Valid Cases**:
The total number of valid distributions is:
\[ 21 - 9 = 12 \]

6. **Account for Distinguishable Apples**:
Since the apples are distinguishable, for each valid distribution of apples among the boys, we can arrange the 8 apples in \( 8! \) ways. Thus, the total number of ways to distribute the apples is:
\[ 12 \times 8! = 12 \times 40320 = 483840 \]

7. **Express in Terms of \( K \) and \( P(7, 3) \)**:
The problem states that the number of ways can be expressed as \( K \cdot P(7, 3) \). We know:
\[ P(7, 3) = 7 \times 6 \times 5 = 210 \]
Therefore, we need to find \( K \) such that:
\[ 483840 = K \cdot 210 \]
Solving for \( K \):
\[ K = \frac{483840}{210} = 2304 \]

### Final Answer:
The value of \( K \) is \( 2304 \).