Competition problem — Combinatorics

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Combinatorics Difficulty 3.9 AMC 10/12 Find the answer China

Suppose set $X = \{1, 2, \dots, 20\}$ . $A$ is a subset of $X$ . The number of the elements of $A$ is at least $2$ and all the elements of $A$ can be arranged as consecutive positive integers. Then the number of such set $A$ is ______.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Each set $A$ satisfying the above conditions can be uniquely determined by its minimum element $a$ and maximum element $b$ , where $a, b \in X$ and $a < b$ . The total number of such ways of taking $(a, b)$ is $C_{20}^2 = 190$ , so the number of such sets $A$ is $190$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.