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Question

The number of ordered pairs (r,k) for which 635Cr=(k23)36Cr+1, where k is an integer, is :
  1. 4
  2. 6
  3. 2
  4. 3


Solution

The correct option is A 4
Using 36Cr+1=36r+1× 35Cr, we get
36r+1× 35Cr×(k23)= 35Cr×6
k23=r+16
k2=r+16+3
kI
r Non-negative integer 0r35
r=5k=±2
r=35k=±3
No. of ordered pairs (r,k)=4

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