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Problem 272

Combinatorics Difficulty 2.3 Multiple choice CEMC Fermat · Canada · 2015

At Barker High School, a total of 36 students are on either the baseball team, the hockey team, or both. If there are 25 students on the baseball team and 19 students on the hockey team, how many students play both sports?

Pick one

Official solutions — 2

Solution 1

The two teams include a total of $25+19=44$ players.

There are exactly 36 students who are at least one team.

Thus, there are $44-36=8$ students who are counted twice.

Therefore, there are 8 students who play both baseball and hockey.

Solution 2

Suppose that there are $x$ students who play both baseball and hockey.

Since there are 25 students who play baseball, then $25-x$ of these play baseball and not hockey.

Since there are 19 students who play hockey, then $19-x$ of these play hockey and not baseball.

Since 36 students play either baseball or hockey or both, then $(25-x)+(19-x)+x = 36$ (The left side is the sum of the numbers of those who play baseball and not hockey, those who play hockey and not baseball, and those who play both.)

Therefore, $44-x = 36$ and so $x = 44-36=8$ .

Thus, 8 students play both baseball and hockey.

Solution 2

Solution 1

The two teams include a total of $25+19=44$ players.

There are exactly 36 students who are at least one team.

Thus, there are $44-36=8$ students who are counted twice.

Therefore, there are 8 students who play both baseball and hockey.

Solution 2

Suppose that there are $x$ students who play both baseball and hockey.

Since there are 25 students who play baseball, then $25-x$ of these play baseball and not hockey.

Since there are 19 students who play hockey, then $19-x$ of these play hockey and not baseball.

Therefore, $44-x = 36$ and so $x = 44-36=8$ .

Thus, 8 students play both baseball and hockey.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.