Assuming that a year consists of 365 days, the probability that a group of n people $\left( n\le 365 \right)$ have all different birthdays is
$\begin{align}
  & a.\dfrac{365}{{{\left( 365 \right)}^{n}}} \\
 & b.\dfrac{n}{{{365}^{n}}} \\
 & c.\dfrac{^{365}{{P}_{n}}}{{{365}^{n}}} \\
 & d.\dfrac{^{n}{{P}_{n}}}{{{365}^{n}}} \\
\end{align}$

Hint: In this question, we will use a simple probability formula and the concept of permutation and combination. $\text{Probability=}\dfrac{\text{Number of favourable cases}}{\text{Total number of cases}}$. If there are n outcomes in an experiment, which is equally likely, mutually exclusive and out of these n outcomes, m cases in favour of happening and event A, then the probability of an event A, will be, $P\left( A \right)=\dfrac{m}{n}=\dfrac{n\left( A \right)}{n\left( S \right)}$. We will use the formula of combination, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and the formula of permutation which is $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.

Complete step by step answer:
Here, we will use the multiplication theorem of combination, each person in the group has 365 possibilities of his birthday, so for n people, the total possibilities of birthdays is equal to ${{\left( 365 \right)}^{n}}$. So, we can write,
Total possibilities of the birthday of n people in a year = ${{\left( 365 \right)}^{n}}$.
We can also write it as, $n\left( S \right)={{\left( 365 \right)}^{n}}$, where n(S) is the total number of probabilities.
Now, in the question it is given that a group of n people should all have different birthdays.
So, we will use the concept of combination and distribution here. Firstly, we choose n different days in a year for n people, that is equal to $^{365}{{C}_{n}}$. Then, n people are chosen for their birthdays, so this is distribution, which is equal to $n!$.
So, total favourable cases for that event = $^{365}{{C}_{n}}\times n!$.
We can also write it as $n\left( A \right){{=}^{365}}{{C}_{n}}\times n!$, where n(A) is the number of favourable cases.
We know that $^{365}{{C}_{n}}=\dfrac{365!}{n!\left( 365-n \right)!}$, so we can write n(A) as,
$\begin{align}
  & n\left( A \right)=\dfrac{365!}{n!\left( 365-n \right)!}\times n! \\
 & n\left( A \right)=\dfrac{365!}{\left( 365-n \right)!} \\
\end{align}$
Now, we know that $^{365}{{P}_{n}}=\dfrac{365!}{\left( 365-n \right)!}$, so we get,
$\begin{align}
  & n\left( A \right)=\dfrac{365!}{\left( 365-n \right)!}{{=}^{365}}{{P}_{n}} \\
 & n\left( A \right){{=}^{365}}{{P}_{n}} \\
\end{align}$
Now, we know that probability can be found out using the formula,
$\text{Probability=}\dfrac{\text{Number of favourable cases}}{\text{Total number of cases}}=\dfrac{n\left( A \right)}{n\left( S \right)}$
So, on substituting the values, we get,
$\text{Probability}=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{^{365}{{P}_{n}}}{{{365}^{n}}}$

Therefore, we get the probability that a group of n people all have different birthdays in a year as $\dfrac{^{365}{{P}_{n}}}{{{365}^{n}}}$. Hence, option (c) is the correct answer.

Note: In this question, we should remember the concept of distribution. Many times, the students forget this and make mistakes. So, the students must remember that we have to permute not only combination, but also distribution.