8 different balls have to be distributed among 3 children so that every child receives at least 1 ball, the total number of ways to do this is:

Options

A. `3^(8)`
B. `""^(8)C_(3)`
C. `8^(3)`
D. `3^(8)-""^(3)C_(1)xx2^(8)+""^(3)C_(2)`

Answer

To solve the problem of distributing 8 different balls among 3 children such that each child receives at least one ball, we can use the principle of inclusion-exclusion. Here is the step-by-step solution:

### Step 1: Calculate Total Distributions Without Restrictions
First, we calculate the total number of ways to distribute 8 different balls to 3 children without any restrictions. Each ball can go to any of the 3 children, so the total number of distributions is given by:
\[ 3^8 \]

### Step 2: Subtract Cases Where At Least One Child Gets No Ball
Next, we need to subtract the cases where at least one child does not receive a ball. We can use the inclusion-exclusion principle for this.

#### Step 2.1: Choose 1 Child to Exclude
If we choose 1 child to exclude, the remaining 2 children can receive the balls. The number of ways to choose 1 child from 3 is given by:
\[ \binom{3}{1} = 3 \]
For the remaining 2 children, the number of ways to distribute the 8 balls is:
\[ 2^8 \]
Thus, the total number of ways where at least one child does not receive a ball is:
\[ \binom{3}{1} \cdot 2^8 = 3 \cdot 256 = 768 \]

### Step 3: Add Back Cases Where Two Children Get No Balls
Now, we have subtracted too much, as we have excluded cases where exactly two children do not receive any balls. We need to add these back.

#### Step 3.1: Choose 2 Children to Exclude
The number of ways to choose 2 children from 3 is:
\[ \binom{3}{2} = 3 \]
In this case, all 8 balls must go to the remaining child, which can be done in:
\[ 1^8 = 1 \]
Thus, the total number of ways where exactly two children do not receive a ball is:
\[ \binom{3}{2} \cdot 1^8 = 3 \cdot 1 = 3 \]

### Step 4: Apply Inclusion-Exclusion Principle
Now we can combine these results using the inclusion-exclusion principle:
\[ \text{Total ways} = 3^8 - \binom{3}{1} \cdot 2^8 + \binom{3}{2} \cdot 1^8 \]
Substituting the values we calculated:
\[ \text{Total ways} = 6561 - 768 + 3 \]
Calculating this gives:
\[ \text{Total ways} = 6561 - 768 + 3 = 5796 \]

### Final Answer
The total number of ways to distribute 8 different balls among 3 children such that each child receives at least one ball is:
\[ \boxed{5796} \]