Question

1. the table shows the number of members of a local chess club who attended each of the three monthly meetings of the club last summer. if no members left or joined the club during the summer, what is the least number of members that the club could have had during the summer?

Explanation:

Step1: Use the principle of inclusion - exclusion

To find the least number of members, assume maximum overlap of attendees.
The maximum number of people who could have attended both June and August is the minimum of \(21\) and \(14\), which is \(14\).
Let \(A\) be the set of June attendees (\(\vert A\vert=21\)), \(B\) be the set of July attendees (\(\vert B\vert = 7\)), and \(C\) be the set of August attendees (\(\vert C\vert=14\)).
The formula for \(\vert A\cup B\cup C\vert=\vert A\vert+\vert B\vert+\vert C\vert-\vert A\cap C\vert-\vert A\cap B\vert-\vert B\cap C\vert+\vert A\cap B\cap C\vert\). To minimize \(\vert A\cup B\cup C\vert\), assume \(\vert A\cap B\vert = 7\) (since \(\vert B\vert = 7\)) and \(\vert B\cap C\vert=7\) (max overlap with \(B\)) and \(\vert A\cap B\cap C\vert = 7\) (max overlap).
\(\vert A\cup B\cup C\vert=21 + 7+14-14 - 7-7+7\)

Step2: Simplify the expression

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Answer:

\(21\)