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This proof of the fact that $n \choose r$ = $n-1 \choose r-1$ + $n-1 \choose r$ is correct but very baffling to me.

Proof: Let X be an n-set, and suppose x is a chosen member of X. The set of all r-subsets of X can be split as follows:

U= r-subsets which contain x.

V= r-subsets which do not contain x.

An r-subset is in U if when x is removed from it we obtain an (r-1)-subset of the (n-1)-set X-{x}. So:

|U|= $n-1 \choose r-1$

On the other hand, an r-subset is in V if it is an r-subset of the (n-1)-set X-{x}. So:

|V|= $n-1 \choose r$.

Addition of U and V gives the result.

But: Surely, |U|= $n-1 \choose r-1$ is only the case when we remove x. How can we remove x without changing the result?

MathFail
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Orm
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  • What do you mean? Having fixed $x\in X$, a subset $S$ of size $n$ must be one of the following (mutually exclusive) types: Either $x\in S$ or $x\not \in S$. If $x\in S$ then the rest of $S$ must be a subset of size $n-1$ chosen from $X-{x}$. If $x\not \in S$ then $S$ must be a subset of size $n$ chosen from $X-{x}$. – lulu Jul 30 '22 at 15:54
  • Note: in the above, I used $n$ where you had used $r$. – lulu Jul 30 '22 at 16:00
  • To get a set in $U$, you choose $x$, then you choose $r-1$ of the remaining $n-1$ elements of $X$. – eyeballfrog Jul 30 '22 at 16:12

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