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Q.
The number of ordered triplets of positive integers which are solutions of the equation $x+y+z=100$ is
Permutations and Combinations
Solution:
The number of triplets of positive integers which are solutions of $x+y+z=100$
$=$ Coefficient of $x^{100}$ in $\left(x+x^{2}+x^{3}+\ldots\right)^{3}$
$=$ Coefficient of $x^{100}$ in $x^{3}(1-x)^{-3}$
$=$ Coefficient of $x^{100}$ in
$x^{3}\left[1+3 x+6 x^{2}+\ldots+\frac{(n+1)(n+2)}{2} x^{n}+\ldots\right] $
$=\frac{(97+1)(97+2)}{2}$
$=49 \times 99=4851 .$