1 vote

At any time, the total number of persons on earth who have shaken hands an odd number of times has to be

The answer provided is even number but cannot understand how

The answer provided is even number but cannot understand how

1 vote

Each handshake involves 2 people.

If the ith person shakes hands $n_{i}$ times, then the sum of the $n_{i}$ be even.

If there were an odd number of odd $n_{i}$ then the sum would be odd.

or proof by parity argument as shown :

**Let E be the set of people who have shaken hands an even number of times.**

**Let O be the set of people who have shaken hands an odd number of times.**

**Let n be the number of times any hand has shaken another.**

Let *n_o* be the number of times a hand from the set *O* has shaken another.

Let *n_e* be the number of times a hand from the set *E* has shaken another.

**Note: n = n_o + n_e**

Each handshake increases *n* by two, hence *n* is even.

*n_e* is a sum of even numbers (by definition) and hence is also even.

Since both *n *and *n_e* are both even, the equation above tells us that *n_o* is even as well.

But *n_o* is a sum of odd numbers (by definition). The only way for *n_o* to be even then, is if the size of *O* is even.