A) 6005
B) 4851
C) 5081
D) None of these
Correct Answer: B
Solution :
The number of triplets of positive integers which are solutions of \[x+y+z=100\]. \[=\]coefficient of \[{{x}^{100}}\] in \[{{(x+{{x}^{2}}+{{x}^{3}}+.....)}^{3}}\] = coefficient of \[{{x}^{100}}\] in \[{{x}^{3}}{{(1-x)}^{-3}}\] = coefficient of \[{{x}^{100}}\] in\[{{x}^{3}}\left( 1+3x+6{{x}^{2}}+....+\frac{(n+1)(n+2)}{2}{{x}^{n}}+..... \right)\] \[=\frac{(97+1)(97+2)}{2}=49\times 99=4851\].
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