If x is a positive integer- then x- x-1- x-2- x-1- x-2- x-3- x-2- x-3- x-4- is equal to

The correct option is A 2x!(x+1)!(x+2)! x! amp; (x+1)! amp; (x+2)! (x+1)! amp; (x+2)! amp; (x+3)! (x+2)! amp; (x+3)! am ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Definition of a Determinant

If x is a p...

Question

If $x$ is a positive integer, then
$∣ ∣ ∣ ∣ ∣ \begin{matrix} x! & (x + 1)! & (x + 2)! (x + 1)! & (x + 2)! & (x + 3)! (x + 2)! & (x + 3)! & (x + 4)! \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

2 x! (x + 1)!

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

2 x! (x + 1)! (x + 2)!

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

2 x! (x + 3)!

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

2 (x + 1)! (x + 2)! (x + 3)!

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

Open in App

Solution

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12

a

4^{x} + 2^{2 x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

I = \int_{2}^{3} \frac{2 x^{5} + x^{4} - 2 x^{3} + 2 x^{2} + 1}{(x^{2} + 1) (x^{4} - 1)} d x

I

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 {2 x}^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12,

a^{2}

Join BYJU'S Learning Program