The total number of ways in which 5 balls of differert colours can be distributed among 3 persons so thai each person gets at least one ball is

Options

A. 75
B. 150
C. 210
D. 243

Explanation

To solve the problem of distributing 5 balls of different colors among 3 persons such that each person gets at least one ball, we can use the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Identify Variables**:
- Let \( n = 5 \) (the number of different colored balls).
- Let \( r = 3 \) (the number of persons).

2. **Use the Inclusion-Exclusion Principle**:
The formula to find the number of ways to distribute \( n \) distinct objects into \( r \) groups such that no group is empty is given by:
\[ r^n - \binom{r}{1}(r-1)^n + \binom{r}{2}(r-2)^n - \binom{r}{3}(r-3)^n + \ldots + (-1)^{r-1}\binom{r}{r-1}(r-(r-1))^n \]

3. **Substitute Values**:
For our case, we substitute \( n = 5 \) and \( r = 3 \):
\[ 3^5 - \binom{3}{1} \cdot 2^5 + \binom{3}{2} \cdot 1^5 - \binom{3}{3} \cdot 0^5 \]

4. **Calculate Each Term**:
- Calculate \( 3^5 \):
\[ 3^5 = 243 \]
- Calculate \( \binom{3}{1} \cdot 2^5 \):
\[ \binom{3}{1} = 3 \quad \text{and} \quad 2^5 = 32 \quad \Rightarrow \quad 3 \cdot 32 = 96 \]
- Calculate \( \binom{3}{2} \cdot 1^5 \):
\[ \binom{3}{2} = 3 \quad \text{and} \quad 1^5 = 1 \quad \Rightarrow \quad 3 \cdot 1 = 3 \]
- Calculate \( \binom{3}{3} \cdot 0^5 \):
\[ \binom{3}{3} = 1 \quad \text{and} \quad 0^5 = 0 \quad \Rightarrow \quad 1 \cdot 0 = 0 \]

5. **Combine All Terms**:
Now we can combine all the terms:
\[ 243 - 96 + 3 - 0 = 243 - 96 + 3 = 150 \]

6. **Final Answer**:
Thus, the total number of ways in which 5 balls of different colors can be distributed among 3 persons such that each person gets at least one ball is:
\[ \boxed{150} \]