Answer : (b) \(\frac{1}{2}\)20C10
(1 + x) 20 = 20C0 + 20C1 x + 20C2 x2 + 20C3 x3 + ..... + 20C19 x19 + 20C20 x20
On putting x = – 1, we get
0 = 20C0 – 20C1 + 20C2 – 20C3 + ..... – 20C19 + 20C20
⇒ 0 = 20C0 – 20C1 + ..... – 20C9 + 20C10 – 20C9 + 20C8 .... + 20C0
(∵ nCr = nCn–r )
