Question

Find the no. of +ve integral solutions of equation x + y + z = 30 such that x > y > z.

• A. 61
• B. 60
• C. 80
• D. 62

Answer 1

Answer: (A)

61

Answer 2

Let y = z + α₁, x = y + α₁ + α₂

Then equation is 3z + 2α₁ + α₂ = 30.

No. of +ve integral solutions of this equation is co-efficient of x³⁰ in

(x³ + x⁶ x⁹ + ...... + x²⁷)

(x² + x⁴ + x⁶ + ...... + x²⁶) (x + x² + ..... + x²⁵).

= $\frac{x^{3}(x^{27} - 1)}{x^{3} - 1}x\frac{x^{2}(x^{26} - 1)}{x^{2} - 1}x\frac{x(x^{25} - 1)}{(x - 1)}$.

Coefficient of x²⁴ in (1 - x²⁷) x (1 - x²⁶) x (1 - x²⁵) (1 - x³)^-1 (1 - x²)^-1 (1 - x)^-1.

Coefficient of x²⁴ in = (1 - x³)^-1 (1 - x²)^-1 (1 - x^-1) since remaining have higher power than 24.

= (1 + x³ + x⁶ + x⁹ .......) (1 + x² + x⁴ + .......) (1 + x + x² ........)

= (1 + 1 + 1 + 1 + 1) + (1 + 1 + 1 + 1) + (1 + 1 + 1 + 1) + (1 + 1 + 1 + 1) + (1 + 1 + 1 + 1) +

Method II: Total no. of +ve integral solution = ²⁹C₂.

No. of solutions in which x = y is 14.

No. of solutions in which y = z is 14

No. of solutions in which z = x is 14

No. of solutions in which x = y = z is 1

No. of solutions in which x ≠ y ≠ z is ²⁹C₂ = 3 x 14 + 2 = 366.

No. of solutions in which x > y > z = $\frac{366}{\angle 3}$ = 61.