How many solutions

$\begin{aligned} S & = \sum_{k=m}^n {n \choose k} {k \choose n} \\ & = \sum_{k=m}^n \frac {n!}{k!(n-k)!} \cdot \frac {k!}{m!(k-m)!} \\ & = \sum_{\color{#3D99F6}k=m}^{\color{#3D99F6}n} \frac {n!}{m!(n-k)!(k-m)!} \\ & = \sum_{\color{#D61F06}k=0}^{\color{#D61F06}{n-m}} \frac {n!}{m!(n-m-k)!(k)!} \\ & = {\color{#3D99F6}\frac {n!}{m!(n-m)!}}\sum_{k=0}^{n-m} \frac {(n-m)!}{k!(n-m-k)!} & \small \color{#3D99F6} \text{Note that }{n \choose k} = \frac {k!}{k!(n-k)!} \\ & = {\color{#3D99F6}{n \choose m}}{\color{#D61F06}\sum_{k=0}^{n-m} {n-m \choose k}} & \small \color{#D61F06} \text{and }\sum_{k=0}^p {p \choose k} = 2^p \\ & = {n \choose m}{\color{#D61F06}2^{n-m}} \end{aligned}$

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$\implies x = \boxed{n-m}$

Let's solve this with a story!

On a planet of $n$ people, $m$ people must be chosen for a council. Each of the non-council members will either be allowed to live or will be killed. In how many ways can this be done?

Approach 1: Choose who lives, then choose the council from the living.

Approach 2: Choose the council, then decide if each of the non-council members lives or is killed.

Both Approach 1 and 2 give all of the possible ways to do this -- and they represent the left-hand and right-hand sides of the equation, respectively. There are 2 choices (lives/killed) for each of $n-m$ people not on the council in Approach 2, so $x=n-m.$

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In Approach 1, we choose $k$ people to live. Now, choose $m$ of those $k$ people to be on the council. Do this for all possible $k$ , noting that we need $k \ge m$ since the council members must be alive. In approach 2, choose $m$ people to be on the council. Now, for each of the remaining $n-m$ people, decide whether they live or are killed.