Question

Find the value of n such that ${ }^{ n } P_{5}=42{ }^{ n } P_{3}, n >4$

• A. $10$
• B. $15$
• C. $12$
• D. $20$

Answer 1

Answer: (A)

We have $^{n}P_{5} = 42 ^{n}P_{3} $
$ \Rightarrow n\left(n - 1\right) \left(n - 2\right) \left(n - 3\right) \left(n - 4\right) = 42 n\left(n - 1\right) \left(n - 2\right) $
$\Rightarrow \left(n-3\right)\left(n-4\right)= 42$
[Since $n > 4$, so $n\left(n-1\right)\left(n-2\right) \ne0$]
$\Rightarrow n^{2} - 7n - 30 = 0$
$\Rightarrow n^{2} - 10n + 3n - 30 = 0$
$ \Rightarrow n= 10$ or $n= -3 $
As n cannot be negative, so $n = 10 $