If $^{9} P_{5} + 5 \cdot^{9} P_{4} =^{10} P_{r}$ , then $' r' =$

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$- 5$

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answer is D.

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Detailed Solution

L.H.S. $=^{9} P_{5} + 5 \cdot^{9} P_{4} = = \frac{9!}{(9 - 5)!} + 5 \frac{9!}{(9 - 4)!} = \frac{9!}{4!} + 5 \frac{9!}{5!}$

$= \frac{9!}{4!} + \frac{5 \cdot 9!}{5 \times 4!} = \frac{9!}{4!} + \frac{9!}{4!}$

$= 2 \cdot \frac{9!}{4!} = \frac{5 \times 2 \times 91}{5 \times 4!} = \frac{10 \times 9!}{5 \times 4!}$

$∴ L.H.S. = \frac{10!}{5!}$

$∴ R.H.S. = \frac{10!}{(10 - r)!}$

$L . H . S . = R . H . S .$ (given)

$\frac{10!}{5!} = \frac{10!}{(10 - r)!}$

$\Rightarrow 10 - r = 5 \Rightarrow r = 5$ .

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