In quick sort, $\frac{n}{10}$ element is chosen as pivot using an algorithm which take time as $n^{2}$.
In worst case the $\frac{n}{10}$ element is either smallest or larget element in the input, then in
that case it will get placed at exteme end.
Then,
$T(n) = T(n-1) + n + n^{2}$
(Here, n is for PARTITION algorithm and $n^{2}$ is for the algorithm used for selection of pivot element)
As $n^{2}$ > $n$, then we can neglect it.
So, $T(n) = T(n-1) + n^{2}$
By substitution, we get,
$T(n) = 1^{2} + 2^{2} +3^{2} + - - - - - - + (n-1)^{2} + n^{2}$
T(n) = $\frac{n(n+1)(2n+1)}{6}$
T(n) = O($n^{3}$)
Hence, Option C is Correct Answer.
NOTE : Here, we have to be careful because question is saying $\frac{n}{10}$ element is chosen as pivot.
If question says $\frac{n}{10}$ smallest or $\frac{n}{10}$ largest element is chosen as pivot then our equation
will change to T(n) = $T(\frac{n}{10}) + T(\frac{9n}{10}) + n + n^{2}$
