Question

Consider the following statements
Statement I If $5{}^4{P}_r=6{}^5{P}_{r-1}$, then $r=8,3$
Statement II If ${}^5{P}_r={}^6{P}_{r-1}$, then $r=5$
Which of the above statements (s) is/are true?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I nor II

Answer 1

Answer: (D)

I. We have, $5{}^4{P}_r=6{}^5{P}_{r-1}$
$\Rightarrow 5\times \frac{4!}{(4-r)!}=6\times \frac{5!}{(5-r+1)!}$
$\Rightarrow \frac{5!}{(4-r)!}=\frac{6\times 5!}{(5-r+1)(5-r)(5-r-1)!}$
$\Rightarrow (6-r)(5-r)=6$
$\Rightarrow r^2-11r+24=0$
$\Rightarrow r^2-8r-3r+24=0$
$\Rightarrow (r-8)(r-3)=0$
$\Rightarrow r=8$
$\Rightarrow r=3$
But $r\ne 8$ as $r\le 4$
$\therefore r=3$
II. We have, ${}^5{P}_r={}^6{P}_{r-1}$
$\frac{5!}{(5-r)!}=\frac{6!}{[6-(r-1)]!}$
$\Rightarrow \frac{5!}{(5-r)!}=\frac{6\times 5!}{(7-r)!}$
$\Rightarrow \frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!}$
$\Rightarrow (7-r)(6-r)=6$
$\Rightarrow 42-13r+r^2=6$
$\Rightarrow r^2-13r+36=0$
$\Rightarrow (r-4)(r-9)=0$
$\Rightarrow r=4,9$
$r=9$ rejected as ${}^5{P}_9$ is not defined.
$\therefore r-4$