Question

The number of positive integral solutions of $xyz = 60$ is

• A. $^{59}C_2$
• B. $^4C_2 \times ^3C_2 \times ^3C_2$
• C. $^4C_3$
• D. $^3C_1 \times ^4C_0 \times ^4C_4$

Hint:

To find positive integral solutions of $x_1 x_2 \dots x_k = N$, factor $N$ into primes. For each prime factor, distribute its powers among the variables using stars and bars. Multiply the counts for all primes.

Answer 1

The Correct Option is B

Step 1: Prime factorization of 60.
$60 = 2^2 \cdot 3^1 \cdot 5^1$

Step 2: Distribute prime factors among $x, y, z$.
Let $x = 2^{a_1}3^{b_1}5^{c_1}, y = 2^{a_2}3^{b_2}5^{c_2}, z = 2^{a_3}3^{b_3}5^{c_3}$. Then:
$a_1 + a_2 + a_3 = 2, b_1 + b_2 + b_3 = 1, c_1 + c_2 + c_3 = 1$

Step 3: Count solutions using stars and bars.
- $a_1 + a_2 + a_3 = 2 \implies {2+3-1 \choose 3-1} = ^4C_2$
- $b_1 + b_2 + b_3 = 1 \implies ^3C_2$
- $c_1 + c_2 + c_3 = 1 \implies ^3C_2$

Step 4: Total number of solutions.
Total = $^4C_2 \times ^3C_2 \times ^3C_2$