In how many ways can 5 prizes be distributed among 4 boys when every boy can take one or more prizes ?

Answer

To solve the problem of distributing 5 prizes among 4 boys where each boy can take one or more prizes, we can use the concept of combinations with repetitions, specifically the "stars and bars" theorem.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to distribute 5 identical prizes among 4 distinct boys. Each boy can receive one or more prizes, meaning no boy can be left without a prize.

2. **Transforming the Problem**: Since each boy must receive at least one prize, we can start by giving one prize to each boy. This ensures that no boy is left empty-handed. After giving out 4 prizes (one to each boy), we have 1 prize left to distribute.

3. **Setting Up the Equation**: Now, we need to distribute the remaining 1 prize among the 4 boys. This can be represented as finding the number of non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 + x_4 = 1 \]
where \(x_i\) represents the number of additional prizes received by boy \(i\).

4. **Applying the Stars and Bars Theorem**: The stars and bars theorem states that the number of ways to distribute \(n\) identical items (stars) into \(r\) distinct groups (bars) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \(n = 1\) (the remaining prize) and \(r = 4\) (the boys). Thus, we need to calculate:
\[ \binom{1 + 4 - 1}{4 - 1} = \binom{4}{3} \]

5. **Calculating the Combination**: Now we compute:
\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 \]

6. **Final Answer**: Therefore, the total number of ways to distribute the 5 prizes among the 4 boys is **4**.