Question

Number of all possible ways of distributing eight identical apples among three persons is

• A. 45
• B. 42
• C. 39
• D. 36

Hint:

Remember the "stars and bars" method for distribution problems. For $n$ identical items and $k$ distinct recipients, imagine the $n$ items as 'stars' (*) and you need $k-1$ 'bars' (|) to divide them into $k$ groups. The total number of arrangements of these $n$ stars and $k-1$ bars is the answer, which is $^{n+k-1}C_{k-1}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
This problem is about distributing identical items into distinct boxes. It is a classic "stars and bars" problem. Since the apples are identical, the order in which a person receives them does not matter; only the final number of apples each person has is important. A person can receive zero apples.
Step 2: Key Formula or Approach
The number of ways to distribute $n$ identical items among $k$ distinct recipients is equivalent to finding the number of non-negative integer solutions to the equation:
\[ x_1 + x_2 + \dots + x_k = n \] where $x_i$ is the number of items received by the $i$-th recipient. The formula for this is given by:
\[ {^{n+k-1}}C_{k-1} \quad \text{or} \quad {^{n+k-1}}C_{n} \] Step 3: Detailed Explanation
In this problem:
- The number of identical items (apples) is $n=8$.
- The number of distinct recipients (persons) is $k=3$.
We need to find the number of non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 = 8 \] where $x_1, x_2, x_3$ are the numbers of apples received by the first, second, and third person, respectively.
Using the stars and bars formula:
Number of ways = ${^{8+3-1}}C_{3-1}$ \[ = {^{10}}C_2 \] Now, we calculate the value of the combination:
\[ {^{10}}C_2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45 \] Step 4: Final Answer
The number of all possible ways to distribute the eight identical apples among three persons is 45.