IF a +b+c = 21 what is the total number of positive integral solutions ?

Options

A. 109
B. 190
C. 901
D. 910

Answer

To find the total number of positive integral solutions for the equation \( a + b + c = 21 \), we can use the combinatorial method known as "stars and bars." Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to find the number of ways to distribute 21 identical items (the total sum) into 3 distinct groups (the variables \( a, b, c \)) such that each group receives at least one item (since they must be positive integers).

### Step 2: Adjust for Positive Integers
Since \( a, b, \) and \( c \) must be positive integers, we can make a substitution. Let:
- \( a' = a - 1 \)
- \( b' = b - 1 \)
- \( c' = c - 1 \)

This transformation ensures that \( a', b', c' \) are non-negative integers (i.e., they can be zero). The equation now becomes:
\[ (a' + 1) + (b' + 1) + (c' + 1) = 21 \]
which simplifies to:
\[ a' + b' + c' = 18 \]

### Step 3: Apply the Stars and Bars Theorem
Now we need to find the number of non-negative integer solutions to the equation \( a' + b' + c' = 18 \). According to the stars and bars theorem, the number of solutions is given by the formula:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the total number of items to distribute (which is 18) and \( r \) is the number of groups (which is 3).

### Step 4: Substitute Values into the Formula
Here, \( n = 18 \) and \( r = 3 \). Therefore, we substitute these values into the formula:
\[ \binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} \]

### Step 5: Calculate the Combination
Now we calculate \( \binom{20}{2} \):
\[ \binom{20}{2} = \frac{20 \times 19}{2 \times 1} = \frac{380}{2} = 190 \]

### Final Answer
Thus, the total number of positive integral solutions to the equation \( a + b + c = 21 \) is **190**.

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