Quantitative Aptitude Ques 917

Question: If $ {}^{n}C _{r}={}^{n}{C _{r-1}} $ and $ {}^{n}P _{r}={}^{n}{P _{r+1}}, $ then the value of $ n $ is

Options:

A) 3

B) 4

C) 2

D) 5

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ {}^{n}C _{r}={}^{n}{C _{r-1}} $

$ \Rightarrow $ $ \frac{n!}{(n-r)!r!}=\frac{n!}{(n-r+1)!(r-1)!} $

$ \Rightarrow $ $ 1=\frac{(n-r)!r(r-1)!}{(n-r+1)(n-r)!(r-1)!}=\frac{r}{(n-r+1)} $

$ \Rightarrow $ $ n-r+1=r $ … (i) Again, $ {}^{n}P _{r}={}^{n}{P _{r+1}} $

$ \Rightarrow $ $ \frac{n!}{(n-r)!}=\frac{n!}{(n-r-1)!} $

$ \Rightarrow $ $ \frac{1}{(n-r)(n-r-1)!}=\frac{1}{(n-r-1)!} $

$ \Rightarrow $ $ n-r=1 $ … (ii) From Eqs. (i) and (ii), we get $ r=2 $

$ \therefore $ $ n=3 $

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