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Interesting combinatorics problem involving bijections

Question 1: How many subsets of \( U=\{ 1,2,...,50\}\) *are there such that their elements' sum exceeds * \(637\)?

Solution: \(1+...+50=1275=638+637\) Thus for each subset \(S\) whose elements sum to more than \(637\), there is another subset (namely \(\{1,...,50\} \setminus S \), the set with all the elements of \(\{1,...,50\}\) minus all the elements in \(S\)). There are \(2^{50}\) subsets of \(\{1,...,50\}\), so there are \(\boxed{2^{49}}\) subsets that satisfy the property.

Question 2: How many subsets of \( U=\{ 1,2,...,n\}\) are there such that their elements' sum exceeds * \(m\) *, with \(m \le \frac{n(n+1)}{2}\)?

This question is significantly harder as the symmetry argument made in the first question won't hold in the majority of cases.

Note by A L
3 years, 9 months ago

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(Link to source of question 1: http://www.imomath.com/index.php?options=238&lmm=1, where bijections are used in the proof more formally). A L · 3 years, 9 months ago

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