# #Combinatorics

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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

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 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.

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i didn't get the solution can u plz explain the solution and question both. Thanx in ADVANCE :)

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