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, so
36r + 1 35Cr = 6 35Crk2 - 3
r + 1 = 6(k2 - 3)
∵ 0 r 35, we need 1 r + 1 36, i.e.
1 6(k2 - 3) 36 196 k2 9
As k Z, k2 = 4 or 9.
For k2 = 4: k = 2 and r + 1 = 6, so r = 5.
For k2 = 9: k = 3 and r + 1 = 36, so r = 35.
Hence S = \(5, 2), (5, -2), (35, 3), (35, -3)\ and n(S) = 4.
Answer: (2)