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  "body": "\n## Proof\n\nWe compute both sides with the [[factorial formula|033KTKKRV6]].\n\n1. Let $k \\le n$. By the factorial formula, $\\binom{n}{k} = \\dfrac{n!}{k!\\,(n-k)!}$.\n2. The number $n - k$ is a natural number with $n - k \\le n$. So the factorial formula also applies to $\\binom{n}{n-k}$.\n3. Since $n - (n - k) = k$, it gives $\\binom{n}{n-k} = \\dfrac{n!}{(n-k)!\\,k!}$.\n4. The denominators in steps 1 and 3 are equal, because multiplication is commutative. Hence $\\binom{n}{k} = \\binom{n}{n-k}$.\n\n## Idea\n\nThe factorial formula treats $k$ and $n - k$ in the same way: they appear only through the product $k!\\,(n-k)!$.\nSwapping them therefore does not change the value.\n",
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