Question

Observe the following Lists

List I	List II
$A{)}^n{C}_r{+}^n{C}_{r-1}=$	$1{)}^{n+1}{P}_r$
$B{)}^{\frac{{}^n{P}_r}{{}^n{P}_{r-1}}=}$	$2)\frac{n-r+1}{r}$
$C{)}^n{P}_r+r{}^n{P}_{r-1}=$	$3)n-r+1$
$D)\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=$	$4)n+r-1$
	$5{)}^{(n+1)}{C}_r$

• A. A-5, B-3, C-2, D-1
• B. A-5, B-3, C-1, D-2
• C. A-2, B-4, C-3, D-1
• D. A-5, B-4, C-2, D-1

Answer 1

The correct option is B A-5, B-3, C-1, D-2

A:

${}^n{C}_r{+}^n{C}_{r-1}=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r+1)!(r-1)!}$

$=\frac{n!}{(n-r+1)!r!}(n-r+1+r)$

$=\frac{(n+1)!}{(n-r+1)!r!}$

${=}^{n+1}{C}_r$

B:

$\frac{{}^n{P}_r}{{}^n{P}_{r-1}}=\frac{n!(n-r+1)!}{n!(n-r)!}$

$=n-r+1$

C:

${}^n{P}_r+r{.}^n{P}_{r-1}=\frac{n!}{(n-r)!}+r.\frac{n!}{(n-r+1)!}$

$=\frac{n!}{(n-r+1)}(n-r+1+r)$

$=\frac{(n+1)!}{(n-r+1)!}$

${=}^{n+1}{P}_r$

D:

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n!(n-r+1)!(r-1)!}{(n-r)!r!\times n!}$

$=\frac{n-r+1}{r}$