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.

#Combinatorics #MathProblem #Math

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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`\sqrt{2}`	$\sqrt{2}$
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Comments

(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 - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$