Question

The number of ordered triplets of non-negative integers which are solutions of the equation x + y + z = 100 is

• A. 6005
• B. 4851
• C. 5081
• D. 5151

Answer 1

The correct option is D

5151

The number of triplets of non-negative integers which are solutions of x + y + z = 100.

= Coefficient of $x^{100}$ in $(1+x+x^2+x^3+........)(1+x+x^2+x^3+...............)(1+x+x^2+x^3+..........)$

= Coefficient of $x^{100}$ in $(1+x+x^2+x^3+..........{)}^3$

= Coefficient of $x^{100}$ in $(1-x{)}^{-3}$

= Coefficient of $x^{100}$ in $(1{+}^3{C}_{1}x{+}^4{C}_2{x}^2+........{+}^{100+2}{C}_{100}{x}^{100}+............)$ = ${}^{100+2}{C}_{100}$ = ${}^{102}{C}_2$

= $\frac{(100+1)(100+2)}{2}$ = 101 × 51 = 5151.

Alternative

x + y + z = 100

The number of triplets of positive integers which are solutions = ${}^{n+r-1}{C}_{r-1}$ where n = total number of

similar things and r = Similar things need to distributed among people/places

${}^{100+3-1}{C}_{3-1}$ = ${}^{102}{C}_2$ = 5151