Question

The number of ordered triplets of positive integers which are solutions of the equation $x+y+z=100$ is

• A. 5081
• B. 6005
• C. 4851
• D. None of these

Answer 1

Answer: (C)

The number of triplets of positive integers which are solutions of $x+y+z=100$
$=$ Coefficient of $x^{100}$ in ${\left(x+x^2+x^3+\dots \right)}^3$
$=$ Coefficient of $x^{100}$ in $x^3(1-x{)}^{-3}$
$=$ Coefficient of $x^{100}$ in
$x^3\left[1+3x+6x^2+\dots +\frac{(n+1)(n+2)}{2}{x}^n+\dots \right]$
$=\frac{(97+1)(97+2)}{2}$
$=49\times 99=4851.$