**Solution:**
The class has 12 children. The first child shakes hands with the other 11 children. The second child has already shaken hands with the first child, and so has to shake hands with only the other 10 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

11 + 10 + ........ + 2 + 1 = 66.

If there were 12 children in the class, then there were 66 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

11 + 10 + ........ + 2 + 1 ?

Indeed, there is! It simply equals 11 × 12 / 2. Can you show why such a formula holds?

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