The number of ways in which a score of 11 can be made from a throw by three persons,each throwing a single die once, is

Options

A. 45
B. 18
C. 27
D. 68

Explanation

To solve the problem of finding the number of ways in which a score of 11 can be made from a throw by three persons, each throwing a single die once, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
Each person (let's call them A, B, and C) throws a die, and we want the total score from their throws to equal 11. Each die can show a number from 1 to 6.

2. **Setting Up the Equation**:
We need to find non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 = 11 \]
where \(1 \leq x_1, x_2, x_3 \leq 6\) (each \(x_i\) represents the score from each die).

3. **Adjusting the Variables**:
To simplify the problem, we can shift the variables to account for the minimum score of 1 on each die:
\[ y_1 = x_1 - 1, \quad y_2 = x_2 - 1, \quad y_3 = x_3 - 1 \]
This gives us:
\[ y_1 + y_2 + y_3 = 8 \]
where \(0 \leq y_1, y_2, y_3 \leq 5\).

4. **Finding All Combinations**:
We can now find combinations of \(y_1, y_2, y_3\) that satisfy the equation \(y_1 + y_2 + y_3 = 8\) while ensuring that none of the \(y_i\) exceeds 5.

5. **Using Generating Functions**:
The generating function for a single die throw (1 to 6) is:
\[ x + x^2 + x^3 + x^4 + x^5 + x^6 = x \frac{1 - x^6}{1 - x} \]
Therefore, for three dice, the generating function becomes:
\[ \left(x \frac{1 - x^6}{1 - x}\right)^3 \]

6. **Finding Coefficient of \(x^{11}\)**:
We need the coefficient of \(x^{11}\) in:
\[ x^3 (1 - x^6)^3 (1 - x)^{-3} \]
This simplifies to finding the coefficient of \(x^8\) in:
\[ (1 - x^6)^3 (1 - x)^{-3} \]

7. **Expanding the Functions**:
- The expansion of \((1 - x^6)^3\) gives:
\[ 1 - 3x^6 + 3x^{12} - x^{18} \]
- The expansion of \((1 - x)^{-3}\) gives:
\[ \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \]

8. **Finding the Coefficient of \(x^8\)**:
- From \(1\): The coefficient of \(x^8\) is \(\binom{8+2}{2} = 45\).
- From \(-3x^6\): The coefficient of \(x^2\) is \(-3 \cdot \binom{2+2}{2} = -3 \cdot 6 = -18\).
- From \(3x^{12}\): No contribution since \(x^{12}\) exceeds \(x^8\).

9. **Calculating the Total**:
\[ 45 - 18 = 27 \]

### Final Answer:
The total number of ways in which a score of 11 can be made from a throw by three persons, each throwing a single die once, is **27**.