Answer 1

We have `""^21"C"_(6"n") = ""^21"C"_(("n"^2 + 5)`
∴ 6n = n² + 5
∴ n² – 6n + 5 = 0
∴ (n – 5)(n – 1) = 0
If n = 5,  `""^21"C"_(6"n") = ""^21"C"_30`
Which is possible,
∴ n = 5 is discarded.
Also `""^21"C"_(6"n") = ""^21"C"_(("n"^2 + 5)`
∴ `""^21"C"_(21 - 6"n") = ""^21"C"_(("n"^2 + 5)`
∴ n² + 6n – 16 = 0
∴ n + 8 = 0 or n – 2 = 0
∴ n = – 8 or n = 2
But n ∈ N
∴ n = 2