If a,b and c are integers and `age1,bge2 and c ge 3`. If `a+b+c=15`, the number of possible solutions of the equation is

Options

A. 55
B. 66
C. 45
D. None of these

Answer

To solve the problem, we need to find the number of integer solutions to the equation \( a + b + c = 15 \) under the constraints \( a \geq 1 \), \( b \geq 2 \), and \( c \geq 3 \).

### Step-by-Step Solution:

1. **Define New Variables**:
Since \( a \), \( b \), and \( c \) have lower bounds, we can redefine them to simplify the equation:
- Let \( a = p + 1 \) where \( p \geq 0 \) (since \( a \geq 1 \))
- Let \( b = q + 2 \) where \( q \geq 0 \) (since \( b \geq 2 \))
- Let \( c = r + 3 \) where \( r \geq 0 \) (since \( c \geq 3 \))

2. **Substitute into the Equation**:
Substitute \( a \), \( b \), and \( c \) into the equation \( a + b + c = 15 \):
\[ (p + 1) + (q + 2) + (r + 3) = 15 \]
Simplifying this gives:
\[ p + q + r + 6 = 15 \]
Therefore, we can rewrite it as:
\[ p + q + r = 15 - 6 = 9 \]

3. **Count the Non-Negative Integer Solutions**:
Now, we need to find the number of non-negative integer solutions to the equation \( p + q + r = 9 \). This can be done using the "stars and bars" theorem.

The formula for the number of non-negative integer solutions to the equation \( x_1 + x_2 + ... + x_k = n \) is given by:
\[ \binom{n + k - 1}{k - 1} \]
In our case, \( n = 9 \) and \( k = 3 \) (for \( p \), \( q \), and \( r \)):
\[ \text{Number of solutions} = \binom{9 + 3 - 1}{3 - 1} = \binom{11}{2} \]

4. **Calculate the Binomial Coefficient**:
Now we calculate \( \binom{11}{2} \):
\[ \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = \frac{110}{2} = 55 \]

5. **Conclusion**:
Therefore, the number of possible solutions of the equation \( a + b + c = 15 \) under the given constraints is \( 55 \).