To solve the problem of distributing 5 different chocolates to 4 children such that each child gets at least one chocolate, we can follow these steps:
### Step 1: Understand the Distribution Requirement
We need to distribute 5 chocolates (let's call them C1, C2, C3, C4, C5) to 4 children (let's call them A, B, C, D) ensuring that each child gets at least one chocolate.
### Step 2: Set Up the Distribution
Since each child must receive at least one chocolate, we can denote the number of chocolates received by each child as \( x_1, x_2, x_3, x_4 \). We know that:
- \( x_1 + x_2 + x_3 + x_4 = 5 \)
- \( x_1, x_2, x_3, x_4 \geq 1 \)
### Step 3: Adjust the Variables
To simplify the equation, we can redefine the variables by letting:
- \( y_1 = x_1 - 1 \)
- \( y_2 = x_2 - 1 \)
- \( y_3 = x_3 - 1 \)
- \( y_4 = x_4 - 1 \)
This transformation gives us:
- \( y_1 + y_2 + y_3 + y_4 = 1 \)
- \( y_1, y_2, y_3, y_4 \geq 0 \)
### Step 4: Count the Non-Negative Solutions
The equation \( y_1 + y_2 + y_3 + y_4 = 1 \) can be solved using the "stars and bars" theorem. The number of non-negative integer solutions is given by:
\[
\text{Number of solutions} = \binom{n+k-1}{k-1}
\]
where \( n \) is the total number of items to distribute (1 in this case) and \( k \) is the number of containers (4 children). Thus:
\[
\text{Number of solutions} = \binom{1 + 4 - 1}{4 - 1} = \binom{4}{3} = 4
\]
### Step 5: Distribute the Chocolates
Now that we know how to distribute the chocolates in terms of counts, we can distribute the chocolates themselves. The distribution of chocolates can happen in different ways based on the counts we derived.
### Step 6: Calculate the Total Arrangements
For each distribution of chocolates (like 1, 1, 1, 2), we can arrange the chocolates in different ways. The total arrangements can be calculated using the multinomial coefficient:
\[
\frac{5!}{1! \cdot 1! \cdot 1! \cdot 2!} = \frac{120}{2} = 60
\]
### Step 7: Combine the Results
Finally, since there are 4 ways to distribute the chocolates in terms of counts and 60 ways to arrange the chocolates for each count distribution, the total number of ways is:
\[
4 \times 60 = 240
\]
### Final Answer
Thus, the total number of ways to distribute 5 different chocolates to 4 children such that each child gets at least one chocolate is **240**.
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