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Q.7. Find r if

(i) $^{5}P_{r}=2\; ^{6}\! P_{r -1}$

Answers (1)

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Given : $^{5}P_{r}=2\; ^{6}\! P_{r -1}$

To find the value of r.

$^{5}P_{r}=2\; ^{6}\! P_{r -1}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-(r-1))!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6\times 5!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}=2\times \frac{6}{(7-r)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{12}{(7-r)\times (6-r)\times (5-r)!}$

$\Rightarrow \, \, \frac{1}{1}= \frac{12}{(7-r)\times (6-r)}$

$\Rightarrow \, \, (7-r)\times (6-r)=12$

$\Rightarrow \, \, 42-6r-7r+r^2=12$

$\Rightarrow \, \, r^2-13r+30=0$

$\Rightarrow \, \, r^2-3r-10r+30=0$

$\Rightarrow \, \, r(r-3)-10(r-3)=0$

$\Rightarrow \, \, (r-3)(r-10)=0$

$\Rightarrow \, \, r=3,10$

We know that

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

where $0\leq r\leq n$

$\therefore 0\leq r\leq 5$

Thus the value of, $r=3$

Posted by

seema garhwal

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