JEE Advanced 2016 (2)

Probability Level 4

∑ r = 0 99 ( 99 ) ! r ! ( 99 − r ) ! − 3 ∑ r = 0 33 ( 99 ) ! ( 3 r ) ! ( 99 − 3 r ) ! = ? \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

None of these 0 2 4 1

Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

Sourabh Jangid
Nov 19, 2016

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Swapnil Vatsal
Dec 23, 2017

For the first term use (1+x)^99. Put x=1. For the 2nd term Use the first expansion again and put x=1 , x=w x=w^2. You will see the connection. W: cube root of unity(complex)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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