# Subset counting problem, creating a bijection

The question is

What is the probability that a subset of $\{1,2,\dots,12\}$ of size 6 contains the number 1?

The brute force method is to count how many subsets 1, which amounts to counting how many subsets of size 5 there are in $\{2,\dots,12\}$, i.e. $\binom{11}{5}$. Hence the answer is $$\frac{\binom{11}{5}}{\binom{12}{6}} = \frac12.$$ But how could one see that there is an immediate bijection between subsets containing 1 and not containing 1?

The bijection you want is just: "map a subset to its complement". Easy to see that, given a subset $S$ and its complement $\overline S$, then $1$ is in exactly one of the two.
One insight is that selecting $6$ out of $12$ items gives probability one-half that the particular number $1$ falls in the selected subset (because selected and leftover subsets are equal in size).
• does this mean that if the subsets were of size 4, there would be a $1/3$ chance of selecting 1? – user369210 Aug 30 '16 at 21:07
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