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