Question

If $ x=\alpha, y=\beta, z=\gamma $ is a positive integral solution of the equation $ x+y+z=12 $, the probability that it is such that $ \alpha<\beta<\gamma $ is

• A. \( \frac{7}{55} \)
• B. \( \frac{12}{55} \)
• C. \( \frac{13}{55} \)
• D. \( \frac{8}{55} \)

Hint:

• Number of positive integral solutions to \(x_1+...+x_k=n\) is \(\binom{n-1}{k-1}\).
• For solutions with \(x < y < z\), list them systematically starting with the smallest possible value for x.
• Alternatively, for distinct solutions, the number of partitions of n into k distinct parts is related to partitions into parts of any size.

Answer 1

The Correct Option is A

We are given that \( x = \alpha, y = \beta, z = \gamma \) are positive integers satisfying the equation \( x + y + z = 12 \), and we are to find the probability that \( \alpha<\beta<\gamma \).

Total number of positive integral solutions:

\[ \text{Number of solutions to } x + y + z = 12 \text{ with } x, y, z \in \mathbb{Z}^+ = \binom{12 - 1}{3 - 1} = \binom{11}{2} = 55 \]

Favorable outcomes:

We need to count the number of integer solutions where \( \alpha<\beta<\gamma \), with all three variables positive integers and \( \alpha + \beta + \gamma = 12 \).

This is a classic case of counting strictly increasing positive integer triplets summing to 12. We find all such ordered triples \((\alpha, \beta, \gamma)\) with \( \alpha<\beta<\gamma \) and \( \alpha + \beta + \gamma = 12 \).

We list them manually or use a generating approach:

• \(1 + 2 + 9 = 12\)
• \(1 + 3 + 8 = 12\)
• \(1 + 4 + 7 = 12\)
• \(1 + 5 + 6 = 12\)
• \(2 + 3 + 7 = 12\)
• \(2 + 4 + 6 = 12\)
• \(3 + 4 + 5 = 12\)

Thus, total number of favorable outcomes = 7

Required Probability:

\[ \frac{\text{Favorable}}{\text{Total}} = \frac{7}{55} \]

Final Answer: \( \boxed{\frac{7}{55}} \)