If 25 identical things be distributed among 5 person then
The number of ways each receives at least one thing but not more than eleven is

Options

A. `.^(24)C_(4)-5.^(12)C_(4)`
B. `.^(24)C_(4)_5`
C. `.^(24)C_(4)-5^(13)C_(4)`
D. None of these

Answer

To solve the problem of distributing 25 identical things among 5 persons such that each person receives at least one thing but not more than eleven, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Distribution**: Since each person must receive at least one thing, we start by giving one thing to each of the 5 persons. This means we distribute 5 things, leaving us with:
\[ 25 - 5 = 20 \text{ things} \]

2. **Setting Up the Equation**: Now, we need to distribute the remaining 20 things among the 5 persons. Let \( x_1, x_2, x_3, x_4, x_5 \) represent the number of additional things received by each person. Our equation now becomes:
\[ x_1 + x_2 + x_3 + x_4 + x_5 = 20 \]
where \( x_i \geq 0 \) for \( i = 1, 2, 3, 4, 5 \).

3. **Applying the Stars and Bars Theorem**: The total number of non-negative integer solutions to the equation \( x_1 + x_2 + x_3 + x_4 + x_5 = 20 \) can be calculated using the stars and bars method:
\[ \text{Number of ways} = \binom{20 + 5 - 1}{5 - 1} = \binom{24}{4} \]

4. **Condition of Not More Than 11**: We need to subtract the cases where any person receives more than 11 things. Suppose one person, say \( x_1 \), receives 12 or more things. We can set \( x_1' = x_1 - 12 \), which gives us:
\[ x_1' + x_2 + x_3 + x_4 + x_5 = 20 - 12 = 8 \]
The number of non-negative integer solutions to this equation is:
\[ \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} \]

5. **Considering All Persons**: Since any of the 5 persons could be the one receiving more than 11 things, we multiply by 5:
\[ 5 \times \binom{12}{4} \]

6. **Final Calculation**: The total number of valid distributions is then:
\[ \text{Total ways} = \binom{24}{4} - 5 \times \binom{12}{4} \]

### Final Answer:
Now we can compute the values:
- Calculate \( \binom{24}{4} \) and \( \binom{12}{4} \):
\[ \binom{24}{4} = \frac{24 \times 23 \times 22 \times 21}{4 \times 3 \times 2 \times 1} = 10626 \]
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]

- Substitute these values into the final equation:
\[ \text{Total ways} = 10626 - 5 \times 495 = 10626 - 2475 = 8151 \]

Thus, the number of ways to distribute 25 identical things among 5 persons such that each receives at least one but not more than eleven is **8151**.