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  "body": "\n## Proof\n\nFor a set $X$ and a natural number $k$, let $C(X, k)$ be the set of [[$k$-combinations|0JZSRWRBZE]] of $X$, the subsets with exactly $k$ elements. We argue by induction on $n = |X|$ that $|C(X, k)| = \\binom{n}{k}$ for every $k$.\n\n1. Let $n = 0$. Then $X = \\varnothing$, and its only subset $\\varnothing$ has $0$ elements.\n2. So $|C(X, 0)| = 1 = \\binom{0}{0}$ and $|C(X, k+1)| = 0 = \\binom{0}{k+1}$, by the definition of the [[binomial coefficient|092GEF38YD]].\n3. Suppose the claim holds for all sets with $n$ elements. Let $|X| = n + 1$, with a [[bijection|0SNQEF7Z53]] $f : \\set{1, \\dots, n+1} \\to X$.\n4. Put $x = f(n+1)$ and $X' = X \\setminus \\set{x}$. The restriction of $f$ to $\\set{1, \\dots, n}$ is a bijection onto $X'$, so $|X'| = n$.\n5. Only $\\varnothing$ has $0$ elements, so $|C(X, 0)| = 1 = \\binom{n+1}{0}$.\n6. Now fix $k \\ge 0$. Split $C(X, k+1)$ into $C_{\\mathrm{out}}$, the sets that omit $x$, and $C_{\\mathrm{in}}$, the sets that contain $x$. These parts are disjoint.\n7. The sets in $C_{\\mathrm{out}}$ are exactly the $(k+1)$-element subsets of $X'$. By the induction hypothesis, $|C_{\\mathrm{out}}| = \\binom{n}{k+1}$.\n8. For $S \\in C_{\\mathrm{in}}$, the set $S \\setminus \\set{x}$ is a subset of $X'$. By the [[complement principle|03QC59M99B]] in $S$, it has $(k+1) - 1 = k$ elements.\n9. For $T \\in C(X', k)$, the sets $T$ and $\\set{x}$ are disjoint, since $x \\notin X'$. By the [[rule of sum|0EE2V3CHFK]], $T \\cup \\set{x}$ has $k + 1$ elements, so it lies in $C_{\\mathrm{in}}$.\n10. The maps $S \\mapsto S \\setminus \\set{x}$ and $T \\mapsto T \\cup \\set{x}$ are mutually inverse. So they are bijections between $C_{\\mathrm{in}}$ and $C(X', k)$.\n11. By the induction hypothesis, $C(X', k)$ has $\\binom{n}{k}$ elements. Composing a numbering of it with $T \\mapsto T \\cup \\set{x}$ numbers $C_{\\mathrm{in}}$. So $|C_{\\mathrm{in}}| = \\binom{n}{k}$, as in the [[bijection principle|0GV2573FP4]].\n12. By the rule of sum, $|C(X, k+1)| = \\binom{n}{k} + \\binom{n}{k+1}$.\n13. By [[Pascal's rule|0YRG241HQQ]], this equals $\\binom{n+1}{k+1}$. Together with step 5, the claim holds for all $k$, and the induction is complete.\n\n## Idea\n\nFix one element $x$. A subset of size $k+1$ either avoids $x$, or contains $x$ together with $k$ other elements.\nThese two cases are counted by the smaller set, and they add up exactly as the defining recursion of $\\binom{n}{k}$ does.\n",
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