Question

The number of non-negative integer solutions of the equation \[ a + b + 2c = 22 \] is:

• A. \(124\)
• B. \(144\)
• C. \(135\)
• D. \(136\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the number of triplets \((a, b, c)\) such that \(a, b, c \geq 0\) and are integers satisfying the linear equation \(a + b + 2c = 22\).
Step 2: Key Formula or Approach:
Since the coefficient of \(c\) is different, we can iterate over possible values of \(c\).
For a fixed value of \(c = k\), the equation becomes \(a + b = 22 - 2k\).
The number of non-negative integer solutions for \(x + y = n\) is given by \(\binom{n+2-1}{2-1} = n+1\).
Step 3: Detailed Explanation:
The variable \(c\) can take integer values from \(0\) to \(11\) (since \(2c \leq 22 \Rightarrow c \leq 11\)).
Case 1: \(c = 0 \Rightarrow a + b = 22\). Number of solutions = \(22 + 1 = 23\).
Case 2: \(c = 1 \Rightarrow a + b = 20\). Number of solutions = \(20 + 1 = 21\).
Case 3: \(c = 2 \Rightarrow a + b = 18\). Number of solutions = \(18 + 1 = 19\).
...
Case 12: \(c = 11 \Rightarrow a + b = 0\). Number of solutions = \(0 + 1 = 1\).
The total number of solutions is the sum of these values:
\[ S = 23 + 21 + 19 + \dots + 1 \]
This is an Arithmetic Progression (AP) with first term \(A = 23\), last term \(L = 1\), and common difference \(d = -2\).
Number of terms \(n = 12\).
Using the sum formula \(S = \frac{n}{2}(A + L)\):
\[ S = \frac{12}{2}(23 + 1) \]
\[ S = 6 \times 24 \]
\[ S = 144 \]
Step 4: Final Answer:
The total number of non-negative integer solutions is 144, which corresponds to option (B).