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Let $G$ be a group and $\omega \left(G\right)$ be the set of element orders of $G$. Let $k\in \omega \left(G\right)$ and $m_k\left(G\right)$ be the number of elements of order $k$ in $G$. Let nse$\left(G\right)=\{m_k\left(G\right):k\in \omega \left(G\right)\}$. Assume $r$ is a prime number and let $G$ be a group such that nse$\left(G\right)=$ nse$\left(S_r\right)$, where $S_r$ is the symmetric group of degree $r$. In this paper we prove that $G\cong S_r$, if $r$ divides the order of $G$ and $r^2$ does not divide it. To get the conclusion we make use of some well-known results on the prime graphs of finite simple groups and their components.