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Q.7. Find r if

(ii) $^{5}P_{r}=^{6\! \! }P_{r-1}$

Answers (1)

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Given : $^{5}P_{r}=^{6\! \! }P_{r-1}$

To find the value of r.

$^{5}P_{r}=^{6\! \! }P_{r-1}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}= \frac{6!}{(6-(r-1))!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=\frac{6\times 5!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{6}{(7-r)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{6}{(7-r)\times (6-r)\times (5-r)!}$

$\Rightarrow \, \, \frac{1}{1}= \frac{6}{(7-r)\times (6-r)}$

$\Rightarrow \, \, (7-r)\times (6-r)=6$

$\Rightarrow \, \, 42-6r-7r+r^2=6$

$\Rightarrow \, \, r^2-13r+36=0$

$\Rightarrow \, \, r^2-4r-9r+36=0$

$\Rightarrow \, \, r(r-4)-9(r-4)=0$

$\Rightarrow \, \, (r-4)(r-9)=0$

$\Rightarrow \, \, r=4,9$

We know that

$^nP_r=\frac{n!}{(n-r)!}$

where $0\leq r\leq n$

$\therefore 0\leq r\leq 5$

Thus, $r=4$

Posted by

seema garhwal

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