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

6005

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4851

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5081

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4581

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Solution

The correct option is B
4851

The number of triplets of positive integers

Which are solutions of $x + y + z$ = 100 = coefficient of $x^{100}$ in

$x^{3} ⟮ 1 + 3 x + 6 x^{2} + . . . . . + \frac{(n + 1) (n + 2)}{2} x^{n} + . . . . . . ⟯$

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

= coefficient of $x^{100}$ in $x^{3} (1 - x)^{- 3}$ = $\frac{(97 + 1) (97 + 2)}{2}$ = $49 \times 99$ = 4851

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