if that is ur answer

take 7 person

then total number of hand shake = $_{2}^{7}\textrm{C}=21$

Now, odd number of people shaking hands odd number of times.

Now what u tell?

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Best answer

Lets consider a graph where each node represent a person and an edge denotes that two person have done handshake. So, our problem reduces to finding the number of nodes in a graph with odd degree and to see if this is even.

Each edge in a graph contribute 2 to the sum of degrees. i.e.,

Sum of degrees in a graph $ = 2e.$

For any vertex in the graph, its degree is either odd or even. So,

Sum of degrees in a graph = Sum of degrees of odd degree vertices + Sum of degrees of even degree vertices

So,

Sum of degrees of odd degree vertices = Sum of degrees in a graph - Sum of degrees of even degree vertices

Here, both terms on RHS are multiple of 2, as first one is $2e$ and sum of even numbers is always even. The difference of two even numbers is also even, and hence sum of degrees of odd degree vertices in a graph is even. We get an even number when we add odd numbers "**even**" number of times and not otherwise. So, number of vertices with odd degree in any graph must be **even**.

0 votes

answer will be even

first thing in order to shake hands odd no. of time we must haven even no. of person

let we have n persons then each will do handshaking with (n-1) persons.

it does not mean that if 1st person will do (n-1) then 2nd will do (n-2) etc(there is no restriction mentioned in question)

so for each person we have n-1 choices and to make n-1 as odd, we must have n as even

first thing in order to shake hands odd no. of time we must haven even no. of person

let we have n persons then each will do handshaking with (n-1) persons.

it does not mean that if 1st person will do (n-1) then 2nd will do (n-2) etc(there is no restriction mentioned in question)

so for each person we have n-1 choices and to make n-1 as odd, we must have n as even