JEE Advanced 2016 (2)

Probability Level 4

r=099(99)!r!(99r)!3r=033(99)!(3r)!(993r)!=?\sum_{r=0}^{99}\frac{(99)!}{r!(99-r)!}-3\sum_{r=0}^{33}\frac{(99)!}{(3r)!(99-3r)!}=?

Try my set JEE ADVANCED 2016

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