Question: $n! = \binom{n}0 D\_n + \binom{n}1 D\_{n-1} + \binom{n}2 D\_{n-2} + ….. +\binom{n}n D\_0 $

$(D\_0 = 1)$

Attempt:

The number of permutations of n distinct objects is n factorial.

$\binom{n}0 D\_n $ means all the objects aren't in their natural place.

$\binom{n}1 D\_{n-1} $ means we fix one object ( choose 1 from n ) then the rest deranged

.
.
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$\binom{n}n D\_0 $ means all the objects are in place

Answer: 
Because the area is proportional
to the square of the radius:
$A = \pi r^2$
or
$r = \sqrt{A/\pi}$,
which means that
$r$ is proportional
to the square root of the area.

Therefore,
if $A$ is replaced by
$A/2$,
$r$ becomes
$\sqrt{(A/2)/\pi}
=\dfrac{\sqrt{A/\pi}}{\sqrt{2}}
$.
You can just consider the formula for area of a circle,
which is
$ \pi\times r^2.$

That's the area of some circle of some radius $r$.
What radius do we have to use in order to get double the area?

Let's make an inspired guess that we want the radius to be $\sqrt2\times r.$
So now we have a new, obviously larger (larger radius) circle with area
$\pi\times \left(\sqrt2\times r \right)^2$.

But let's see what that's equal to:

\begin{align}
 \pi\times \left(\sqrt2\times r \right)^2
&= \pi \times \left(\sqrt2\right)^2 \times r^2 && \text{because $(ab)^2 = a^2b^2$}\\
&= \pi \times 2 \times r^2 \\
&= 2\times\left(\pi \times r^2\right) && \text{associativity and commutativity of multiplication}\\
\end{align}

So before we had $\pi\times r^2,$ and now we have
$2\times\left(\pi \times r^2\right)$,
that is, exactly twice as much area, where $r$ is still the same number.

So to double the area, we just have to multiply the radius by $\sqrt2.$

The diameter is just exactly twice the radius, so when you multiply the radius by $\sqrt2$ you also multiply the diameter by $\sqrt2.$

So to double the area, multiply the diameter by $\sqrt2.$

If you want to get half the area, just suppose you started with the larger circle; now you have to divide its diameter by $\sqrt2$ to get the diameter of the smaller circle, which has half the area.