a² – a = 2 + 4 ( ⁿC_{x =} ⁿC_y ⇒ n = x + y or x = y )

a² – a – 6 = 0

a² – 3a – 2a – 6 = 0

a(a – 3) + 2(a – 3) = 0

(a + 2) (a – 3) = 0

a = -2 or a = 3

But, a = -2 is not possible since it’s negative.

So, a = 3.