In how many ways we can give 5 apples, 4 mangoes and 3 oranges (fruits of same species are similar) to three persons If each may receive none, one or more?

Answer

To solve the problem of distributing 5 apples, 4 mangoes, and 3 oranges to three persons, we can use the combinatorial method known as "stars and bars." This method is useful for distributing identical objects into distinct groups.

### Step-by-Step Solution:

1. **Distributing Apples:**
We need to distribute 5 identical apples among 3 persons. Let \( a_1, a_2, a_3 \) be the number of apples received by persons 1, 2, and 3 respectively. We need to solve the equation:
\[ a_1 + a_2 + a_3 = 5 \]
Using the stars and bars theorem, the number of non-negative integer solutions is given by:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the number of identical items (5 apples) and \( r \) is the number of persons (3). Thus, we have:
\[ \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} \]
Calculating this gives:
\[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]

2. **Distributing Mangoes:**
Next, we distribute 4 identical mangoes among 3 persons. Let \( m_1, m_2, m_3 \) be the number of mangoes received by persons 1, 2, and 3 respectively. We need to solve:
\[ m_1 + m_2 + m_3 = 4 \]
Again using the stars and bars theorem:
\[ \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} \]
Calculating this gives:
\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]

3. **Distributing Oranges:**
Finally, we distribute 3 identical oranges among 3 persons. Let \( o_1, o_2, o_3 \) be the number of oranges received by persons 1, 2, and 3 respectively. We need to solve:
\[ o_1 + o_2 + o_3 = 3 \]
Using the stars and bars theorem again:
\[ \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} \]
Calculating this gives:
\[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]

4. **Calculating Total Ways:**
Since the distributions of apples, mangoes, and oranges are independent events, we multiply the number of ways to distribute each fruit:
\[ \text{Total Ways} = 21 \times 15 \times 10 \]
Calculating this gives:
\[ 21 \times 15 = 315 \]
\[ 315 \times 10 = 3150 \]

Thus, the total number of ways to distribute 5 apples, 4 mangoes, and 3 oranges among three persons is **3150**.