Q.

The number of elements in the set S={(r,k):k∈Z and Cr+136=6(Cr35)(k2–3)}, is:          [2026]

1 2  
2 4  
3 8  
4 16  

Ans.

(2)

Cr+136=6Cr35(k2–3)

⇒ Cr+136Cr35=6k2–3

⇒ 36!(r+1)!×(35–r)!35!r!(35–r)!=6k2–3

⇒ 36r+1=6k2–3 ⇒ 6(k2–3)=r+1

⇒ 6k2–r=19 ⇒ r=6k2–19

Now, Cr35 is defined, so 0≤6k2≤35

⇒ 0≤k2–19≤35 ⇒ 19≤6k2≤54

⇒ 196≤6k2≤9

⇒ 3.1≤k2≤9 ⇒ k∈{–3,–2,2,3} 

⇒ r∈{35,5}

So, (r, k) = {(35, –3), (35,3), (5, –2), (5, 2)}

  Number of elements = 4.