**Solution:**
The class has 15 children. The first child shakes hands with the other 14 children. The second child has already shaken hands with the first child, and so has to shake hands with only the other 13 children. In this manner, the second-last child has to shake hands with only one child, and the last child has already met all the children. Thus, the number of handshakes is

14 + 13 + ........ + 2 + 1 = 105.

If there were 15 children in the class, then there were 105 total handshakes.

**Food for thought:**
It is obviously assumed that each child shakes hands with every other child once and only once.

More importantly, is there a quick way to add

14 + 13 + ........ + 2 + 1 ?

Indeed, there is! It simply equals 14 × 15 / 2. Can you show why such a formula holds?

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