The value of 12! is closest to:

A. (10^6)
B. 3(10^7)
C. 5(10^8)
D. 7(10^9)
E. 9(10^11)

The OA is C.

How can I determine this? Can eny expert help me with this PS question please? Thanks.

Hi LUANDATO,
Let's take a look at your question.

We are asked to estimate the value of 12!.

$$12!\ =\ 12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1$$
Let's now make groups of products that are aproximately equal to 10 or 100,
$$=\ \left(12\times8\right)\times\left(11\times9\right)\times\left(10\times1\right)\times\left(7\times6\times2\right)\times\left(4\times3\right)\times5$$
$$=\ \left(96\right)\times\left(99\right)\times\left(10\right)\times\left(84\right)\times\left(12\right)\times5$$
$$\approx\left(100\right)\times\left(100\right)\times\left(10\right)\times\left(100\right)\times\left(10\right)\times5$$
$$\approx\left(10^2\right)\times\left(10^2\right)\times\left(10^1\right)\times\left(10^2\right)\times\left(10^1\right)\times5$$
$$\approx\left(10^{2+2+1+2+1}\right)\times5$$
$$\approx\left(10^8\right)\times5$$
$$\approx5\left(10^8\right)$$

Therefore, Option D is correct.

Hope this helps.
I am available if you'd like any follow up.