0

We consider a relation : " a person X shakes hand with Y" . Obviously if X shakes hand with Y , then Y shakes hand with X . In a gathering of 99 persons, one of following statements is always true, considering 0 to be even number which one is it ?

  • A) There is exactly one person who shakes hand exactly with an odd number of persons

  • B) There's exactly one person who shakes hand exactly with an even number of persons

  • C) There are even number of persons who shakes hands with even number of person

  • D) none

I could't understand why 0 is considered even here, and can someone tell me how to do it.

JMoravitz
  • 79,518
thebeatle
  • 193
  • 1
    What have you tried? Hint..it's easy enough in explicit cases. Suppose, say, that no hands are shaken at all. 0 is generally considered even, as 0 = 2* 0. – lulu Jul 15 '15 at 03:44
  • As for "why $0$ is considered even", read the answers at http://math.stackexchange.com/questions/15556/is-zero-odd-or-even – JMoravitz Jul 15 '15 at 04:43

2 Answers2

2

There are two results that are commonly referred to as "the hand-shaking lemma" which I will state here as a theorem and a corollary.

Theorem: Given any simple graph, $G(V,E)$, you have $$\sum\limits_{v\in V} d(v)=2|E|$$

Corollary: In any simple graph, there must be an even number of vertices of odd degree.

Here, $d(v)$ means "the degree of vertex $v$", i.e. the number of edges that $v$ is incident to. If you consider each person at your handshaking party a vertex, and each handshake as an edge, you can treat the situation as a graph and use these results.

Now, look at each of these statements a bit closer:

  • A: Is it always true that there is exactly one person who has shaken an odd number of hands? Can you have a situation where either zero or more than one person shakes an odd number of hands?

What about if noone shakes any hands at all?

  • B: Is it always true that there is exactly one person who shakes hands with an even number of persons? Can you have a situation where either zero or more than one person shakes an even number of hands?

What about if noone shakes any hands at all?

  • C: Is it always true that there are an even number of people who shakes hands with an even number of other people?

What does that say about the number of people who shake an odd number of hands? Does the corollary above have anything to say about this?

A more interesting question is which of these are possibly true versus never true (as opposed to always true).

A: refer to the corollary, C: refer to the corollary, B: Can you come up with a way to have 98 people all shaking an odd number of hands with one person shaking no hands?

Community
  • 1
JMoravitz
  • 79,518
1

In our graph construction, the sum of degree of all vertices will be 2E, E being th enumber of handshakes. Because every handshake increases the total degree by 2.
Now, let's start with the options.

  1. There is exactly one person who shakes hand with an odd number of persons.
    If There is exactly one person with an odd number of handshakes, the degree of his vertex will be odd. So, for being the sum of degree to be even, there must be another vertex(or, person) who has done odd number of handshakes. So,option 1 is false.
  2. There is exactly one person who shakes hand with an even number of persons.
    So, the number of person he shakes hand with, will be an even number, let's say 2n. But we had total 99 people. So, There are 2n people who shakes hands an odd number of times. Also, 99-1-2n = even number of persons are there, each of them shakes hand an odd number of times. So, sum of degree is even constraint satisfies.
  3. There are even number of persons who shake hand even number of times.
    This is not possible. 2n being the number of even number of persons, 99-2n = odd. So, we have odd number of persons who shake hands odd number of times. This clearly violates that sum of degree is an even number.

So, B is the answer in your case.
O is considered even because if nobody shakes hands with nobody, we have 2E = 2*0 = 0 satisfied.

Nishat
  • 41