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: 
The area $A$ of a circle of diameter $d$ is
$$A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4}$$

Let's say we have two diameters $d\_1$ and $d\_2$, and their corresponding areas $A\_1$ and $A\_2$:
$$A\_1 = \frac{\pi d\_1^2}{4}, \quad A\_2 = \frac{\pi d\_2^2}{4}$$
Let's see what the diameters are if $A\_2 = 2 A\_1 \gt 0$ (so $d\_1 \gt 0$ and $d\_2 \gt 0$ too):
$$\begin{aligned}
\frac{\pi d\_2^2}{4} &= 2 \frac{\pi d\_1^2}{4} \\
d\_2^2 &= 2 d\_1^2 \\
\sqrt{d\_2^2} &= \sqrt{2 d\_1^2} \\
d\_2 &= \sqrt{2} d\_1 \\
\end{aligned}$$
Above, we are allowed to take the square root because both sides are nonnegative, since $d\_1 \gt 0$ and $d\_2 \gt 0$.

In other words, the square root comes from the fact that the area is proportional to the square of the diameter; and the square root of 2 comes from the fact that we are comparing areas whose ratio is 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.