Question

If ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_x , then x =
(a) r
(b) r − 1
(c) n
(d) r + 1

Answer 1

(d) r + 1

$n_{C_r}+n_{C_{r+1}}=n+1_{C_x}$ [Given]

We have:
${}^n{C}_r{+}^n{C}_{r+1}={}^{n+1}{C}_x$ [∵${}^n{C}_r{+}^n{C}_{r-1}={}^{n+1}{C}_r$]
$\Rightarrow {}^{n+1}{C}_{r+1}={}^{n+1}{C}_x$
$\Rightarrow r+1=x$ [∵${}^n{C}_x{=}^n{C}_y\Rightarrow n=x+y\mathrm{or}x=y$]