Answer
The required solution is $1$.
Work Step by Step
We know that the representation $_{n}{{C}_{r}}$ counts the number of assortments of n items taken r at a time.
And the number of assortment of n items taken r at a time can be evaluated as:
$_{n}{{C}_{r}}=\frac{n!}{\left( n-r \right)!r!}$
And the provided expression is $_{4}{{C}_{4}}$.
Here, $ n=4,r=4$.
Put the value of n, r in the above formula. Then:
$\begin{align}
& _{4}{{C}_{4}}=\frac{4!}{\left( 4-4 \right)!4!} \\
& =\frac{4!}{0!4!} \\
& =\frac{1}{0!}
\end{align}$
Since, $0!=1$
So, $\begin{align}
& \frac{1}{0!}=\frac{1}{1} \\
& =1
\end{align}$
Hence, $_{4}{{C}_{4}}=1$
