Question: The number of ordered triplets of positive integers which are solutions of the equation $ x+y+z=100 $ is
Options:
A) 6005
B) 4851
C) 5081
D) None of these
Show Answer
Answer:
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}( 1+3x+6x^{2}+….+\frac{(n+1)(n+2)}{2}x^{n}+….. ) $ $ =\frac{(97+1)(97+2)}{2}=49\times 99=4851 $ .