Question

Oliver writes $4!-2!$ as $(4-2)!,$ is he right in his approach?

• A. Yes
• B. No

Answer 1

The correct option is B No

Let's find the values of $4!-2!$ and $(4-2)!$ one-by-one:

$\bullet$ $4!-2!$

$4!-2!=4\times 3\times 2\times 1-2\times 1$

Taking out the common part $(2\times 1)$
$\Rightarrow 4!-2!=(4\times 3-1)\times (2\times 1)$
$\Rightarrow 4!-2!=(12-1)\times \left(2\right)$

$\Rightarrow 4!-2!=\left(11\right)\times \left(2\right)=22$

$\bullet$ $(4-2)!$

$(4-2)!=2!=2\times 1=2$

$\because 22\ne 2$
$\Rightarrow 4!-2!\ne (4-2)!$

Hence, Alex is not right in his approach of writting $4!-2!$ as $(4-2)!$

Therefore, option (b.) is the correct choice.