If (n + 1)! = 6 [(n - 1)!], find n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 2
$(n + 1)! = 6 [(n - 1)!]$

$\Rightarrow (n + 1) . n . [(n - 1)!] = 6 [(n - 1)!]$

$\Rightarrow n^{2} + n = 6 \Rightarrow n^{2} + n - 6 = 0$

$\Rightarrow (n - 2) (n + 3) = 0$

$∴$ Either n - 2 = 0 or n + 3 = 0

$\Rightarrow n = 2 o r n = - 3$
n being natural number, so
$n \neq - 3$

$∴ n = 2$

Suggest Corrections

Similar questions

^{n} P_{3} :^{n} P_{6} = 1 : 210,

n

If 5 P(4,n)= 6 P (5,n-1), find n.

^{n} P_{3} :^{n} P_{6} = 1 : 210

n

n

\frac{9^{n + 1} + 4^{n + 1}}{9^{n} + 4^{n}} = 6

Join BYJU'S Learning Program