1. Tardigrade
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  3. Mathematics
  4. If nPr = nPr + 1 and nCr = nC-1 , then the values of n and r are

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Q. If ${^{n}P_r} = {^{n}P_{r + 1}} $ and $^nC_r = {^{n}C_{-1}}$ , then the values of n and r are

Permutations and Combinations

Solution:

We have, ${^{n}P_r} = {^{n}P_{r+1}}$
$\Rightarrow \frac{n!}{\left(n-r\right)!} = \frac{n!}{\left(n-r-1\right)!} \Rightarrow \frac{1}{\left(n-r\right)} = 1 $
or $n -r = 1 $ ....(1)
Also, $^{n}C_{r} = ^{n}C_{r-1} \Rightarrow r+r-1 = n$
$ \Rightarrow 2r -n = 1 .....(2) $
Solving (1) and (2), we get r = 2 and n = 3