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When do same-group and mixed-group handshakes match?

Count each pair once

Within groups of sizes $b$ and $g$ , the pair count is $b (b - 1) /2 + g (g - 1) /2$ . Across the groups it is $b g$ . Equating the two counts simplifies to $(b - g)^{2} = b + g$ .

Problem

A class has 30 children. On a day when some are absent, every pair of children present shakes hands once. The number of handshakes between children of the same sex equals the number between boys and girls. What is the fewest number absent? If attendance is as large as possible, how many girls could be present?

Answer

At least

5

are absent. There can be

10

15

girls present.

The two possible group sizes are 10 and 15, in either order.

Step-by-step solution

1. Write both counts

If $b$ boys and $g$ girls attend, same-group handshakes total $b (b - 1) /2 + g (g - 1) /2$ . Mixed handshakes total $b g$ .

2. Simplify the equality

Multiplying by 2 and rearranging gives $b^{2} - 2 b g + g^{2} = b + g$ , or $(b - g)^{2} = b + g$ . Attendance must therefore be a perfect square.

3. Use the class limit

The largest square no greater than 30 is 25, so at least $30 - 25 = 5$ are absent. With $b + g = 25$ and $∣ b - g ∣ = 5$ , the groups have sizes 10 and 15. Either size may be the number of girls.

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