Correct option: (c) 2n-1-1
Explanation:
given: n–1c1 + n–1c2 + --- n–1cn–1
we know
nc0 + nc1 + nc2 + nc3 + ------ ncn = 2n
Replacing n by(n – 1), we get
n–1c0 + n–1c1 + n–1c2 + ----- n–1cn–1 = 2n–1
∴ n–1c1 + n–1c2 + ---- n–1cn–1 = 2n–1 – n–1c0
= 2n–1 – 1