Distribution of $n\choose x$

Question

If $X$ follows a Binomial distribution with parameters $n$ and $p$, what can we say on the distribution of $n\choose x$ and $x\choose n$ ? ($x$ is realization of $X$).

Answer 1

These questions can be solved by using algebraic manipulations to transform the probability mass function of $X$ to that of $\binom{n}{X}$ and $\binom{X}{n}$. For the first case, if $n$ is odd then $x\mapsto\binom{n}{x}$ is a two-to-one function of $x$ where $\binom{n}{x}=\binom{n}{n-x}$. Therefore we can say that (for even $n$) \begin{align} \Pr(\binom{n}{X}=\binom{n}{x_0})&=\Pr(X=x_0\;\text{or}\; X=n-x_0)\\ &=\Pr(X=x_0)+\Pr(X=n-x_0)\\ &=\binom{n}{x_0}p^{x_0}(1-p)^{n-x_0}+\binom{n}{n-x_0}p^{n-x_0}(1-p)^{x_0}\\ &=\binom{n}{x_0}\left[ p^{x_0}(1-p)^{n-x_0}+p^{n-x_0}(1-p)^{x_0} \right]. \end{align} If $n$ is even, then $x\mapsto\binom{n}{x}$ is not two-to-one because there is only one $x$ for which $\binom{n}{x}=\binom{n}{n/2}$, in other words $x=n/2=n-x$. Therefore, if $n$ is even then \begin{align} \Pr(\binom{n}{X}=\binom{n}{x_0})&=\begin{cases}\Pr(X=x_0\;\text{or}\;X=n-x_0),& x_0\neq n/2\\ \Pr(X=n/2), & x_0=n/2 \end{cases}\\ &=\begin{cases}\binom{n}{x_0}\left[ p^{x_0}(1-p)^{n-x_0}+p^{n-x_0}(1-p)^{x_0} \right],& x_0\neq n/2 \\ \binom{n}{n/2}p^{n/2}(1-p)^{n/2}, & x_0=n/2. \end{cases} \end{align} For the second part, the range of $X$ is $\{0,\dotsc,n\}$ and the typical definition of $\binom{x}{n}$ is zero for $n>x$, so the possible values that $\binom{X}{n}$ will take are $0$ for $X<n$ and $\binom{n}{n}=1$ for $X=n$. Formally, \begin{align} \Pr(\binom{X}{n}=i)&=\begin{cases} \Pr(X<n),& i=0\\ \Pr(X=n), & i=\binom{n}{n}\end{cases}\\ &=\begin{cases}1-\Pr(X=n), & i=0\\ \Pr(X=n), & i=1\end{cases}\\ &=\begin{cases}1-p^n , & i=0\\ p^n, & i=1.\end{cases} \end{align} The distribution for $\binom{n}{X}$ does not have a name to the best of my knowledge, whereas the distribution of $\binom{X}{n}$ is called Bernoulli($p^n$).