Answer the questions based on the following information. There are four integers a,b,c such that a+b+c=p.
If p=30 and `0le(a,b,c)le15`, find the number of solutions of the given equation.

Options

A. 136
B. 2456
C. 5226
D. 3436

Answer

To solve the problem of finding the number of integer solutions for the equation \( a + b + c = 30 \) under the constraints \( 0 \leq a, b, c \leq 15 \), we can follow these steps:

### Step 1: Rewrite the Variables
Since \( a, b, c \) must be less than or equal to 15, we can introduce new variables to simplify our calculations. Let:
- \( x_1 = 15 - a \)
- \( x_2 = 15 - b \)
- \( x_3 = 15 - c \)

This implies:
- \( a = 15 - x_1 \)
- \( b = 15 - x_2 \)
- \( c = 15 - x_3 \)

### Step 2: Substitute into the Equation
Substituting these new variables into the original equation gives us:
\[ (15 - x_1) + (15 - x_2) + (15 - x_3) = 30 \]
This simplifies to:
\[ 45 - (x_1 + x_2 + x_3) = 30 \]
Rearranging this, we find:
\[ x_1 + x_2 + x_3 = 15 \]

### Step 3: Set Up the Non-Negative Integer Solutions
Now, we need to find the number of non-negative integer solutions to the equation \( x_1 + x_2 + x_3 = 15 \). The number of solutions to the equation \( x_1 + x_2 + x_3 = n \) in non-negative integers is given by the "stars and bars" theorem, which states that the number of solutions is:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the total we want (15 in this case) and \( r \) is the number of variables (3 here).

### Step 4: Apply the Formula
In our case, \( n = 15 \) and \( r = 3 \). Thus, we need to calculate:
\[ \binom{15 + 3 - 1}{3 - 1} = \binom{17}{2} \]

### Step 5: Calculate \( \binom{17}{2} \)
Calculating \( \binom{17}{2} \):
\[ \binom{17}{2} = \frac{17 \times 16}{2 \times 1} = \frac{272}{2} = 136 \]

### Final Answer
Thus, the number of solutions to the equation \( a + b + c = 30 \) with the constraints \( 0 \leq a, b, c \leq 15 \) is **136**.

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