Permutations And Combinations Question 249

Question: The value of $ ^{n}P _{r} $ is equal to [IIT 1971; MP PET 1993]

Options:

A) $ ^{n-1}P _{r}+r{{}^{n-1}}{P _{r-1}} $

B) $ n.{{\ }^{n-1}}P _{r}{{+}^{n-1}}{P _{r-1}} $

C) $ n{{(}^{n-1}}P _{r}{{+}^{n-1}}{P _{r-1}}) $

D) $ ^{n-1}{P _{r-1}}{{+}^{n-1}}P _{r} $

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Answer:

Correct Answer: A

Solution:

  • $ ^{n-1}P _{r}+r{{.}^{n-1}}{P _{r-1}} $ $ =\frac{(n-1)!}{(n-1-r)!}+r\frac{(n-1)!}{(n-r)!} $ $ ( \because {{}^{n}}P _{r}=\frac{n!}{(n-r)!} ) $ = $ \frac{(n-1)!}{(n-1-r)!}{ 1+r.\frac{1}{n-r} } $ = $ \frac{(n-1)!}{(n-1-r)!(n-r)!}( \frac{n}{n-r} )=\frac{n!}{(n-r)!}={{}^{n}}P _{r} $ . Aliter: We know that $ ^{n-1}C _{r}+{{}^{n-1}}{C _{r-1}}={{}^{n}}C _{r} $
    Þ $ \frac{^{n-1}P _{r}}{r!}+\frac{^{n-1}{P _{r-1}}}{(r-1)!}=\frac{^{n}P _{r}}{r!} $
    Þ $ ^{n-1}P _{r}+r.{{}^{n-1}}{P _{r-1}}={{}^{n}}P _{r} $ .