The number of elements in the set S= \(r,k) : k Z and 36Cr+1= 6 (35Cr ) (k2-3 ) \, is :
- 2
- 4
- 8
- 16
Answer
(2) 4
Solution
Using nCr= nr n-1Cr-1: 36Cr+1= 36r+1 35Cr
∴ 36r+1 35Cr= 6 35Crk2-3
r+1=6(k2-3)
∵ 0 r 35 1 r+1 36
1 6(k2-3) 36 196 k2 9
k Z k2=4 or 9
k2=4: k= 2, r+1=6 r=5
k2=9: k= 3, r+1=36 r=35
S=\(5,2),(5,-2),(35,3),(35,-3)\ n(S)=4
Answer: (2)