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, $5{ }^4 P_r=6{ }^5 P_{r-1}$
$\Rightarrow 5 \times \frac{4 !}{(4-r) !}=6 \times \frac{5 !}{(5-r+1) !}$
$\Rightarrow \frac{5 !}{(4-r) !}=\frac{6 \times 5 !}{(5-r+1)(5-r)(5-r-1) !}$
$\Rightarrow (6-r)(5-r)=6$
$\Rightarrow r^2-11 r+24=0$
$\Rightarrow r^2-8 r-3 r+24=0$
$\Rightarrow (r-8)(r-3)=0$
$\Rightarrow r=8$
$\Rightarrow r=3$
But $r \neq 8$ as $r \leq 4$
$\therefore r=3$
II. We have, ${ }^5 P_r={ }^6 P_{r-1}$
$\frac{5 !}{(5-r) !}=\frac{6 !}{[6-(r-1)] !}$
$\Rightarrow \frac{5 !}{(5-r) !}=\frac{6 \times 5 !}{(7-r) !}$
$\Rightarrow \frac{1}{(5-r) !}=\frac{6}{(7-r)(6-r)(5-r) !}$
$\Rightarrow (7-r)(6-r)=6$
$\Rightarrow 42-13 r+r^2=6$
$\Rightarrow r^2-13 r+36=0$
$\Rightarrow (r-4)(r-9) =0$
$\Rightarrow r =4,9$
$r=9$ rejected as ${ }^5 P_9$ is not defined.
$\therefore r-4$