If a,b, and c are positive integers such that a+b+c`le8`, the number of possible values of the ordered triplet (a,b,c) is

Options

A. 84
B. 56
C. 83
D. None of these

Answer

To solve the problem of finding the number of ordered triplets (a, b, c) such that a + b + c ≤ 8, where a, b, and c are positive integers, we can follow these steps:

### Step 1: Transform the variables
Since a, b, and c are positive integers, we can express them in terms of new variables that are non-negative integers. Let:
- \( a = a_1 + 1 \)
- \( b = b_1 + 1 \)
- \( c = c_1 + 1 \)

Here, \( a_1, b_1, c_1 \) are non-negative integers (i.e., \( a_1, b_1, c_1 \geq 0 \)).

### Step 2: Rewrite the inequality
Substituting these expressions into the inequality gives:
\[ (a_1 + 1) + (b_1 + 1) + (c_1 + 1) \leq 8 \]
This simplifies to:
\[ a_1 + b_1 + c_1 + 3 \leq 8 \]
or equivalently:
\[ a_1 + b_1 + c_1 \leq 5 \]

### Step 3: Introduce a new variable
To convert the inequality into an equation, we introduce a new variable \( n \):
\[ a_1 + b_1 + c_1 + n = 5 \]
where \( n \) is a non-negative integer that accounts for the difference between the left-hand side and 5.

### Step 4: Count the non-negative integer solutions
Now we need to find the number of non-negative integer solutions to the equation:
\[ a_1 + b_1 + c_1 + n = 5 \]
This is a classic problem that can be solved using the "stars and bars" theorem. The formula for the number of solutions in non-negative integers to the equation \( x_1 + x_2 + ... + x_k = N \) is given by:
\[ \binom{N+k-1}{k-1} \]
In our case, \( N = 5 \) and \( k = 4 \) (since we have \( a_1, b_1, c_1, n \)), so we calculate:
\[ \binom{5 + 4 - 1}{4 - 1} = \binom{8}{3} \]

### Step 5: Calculate the binomial coefficient
Now we compute \( \binom{8}{3} \):
\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

### Conclusion
The total number of possible values of the ordered triplet (a, b, c) such that \( a + b + c \leq 8 \) is **56**.

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