**Solution:**
The class has 10 children. The first child shakes hands with the other 9 children. The second child has already shaken hands with the first child, and so has to shake hands with only the other 8 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

9 + 8 + ........ + 2 + 1 = 45.

If there were 10 children in the class, then there were 45 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

9 + 8 + ........ + 2 + 1 ?

Indeed, there is! It simply equals 9 × 10 / 2. Can you show why such a formula holds?

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