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In a party people shake hands with one another (not necessarily every one with every one else). A) show that two persons shake hands the same number of times. b) show that the number of people who shake hands an odd number of times is even.

Soumya Sharma
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    Edit your question to show what you've tried and where you are stuck. Can you work out some small examples? What might happen if there were just two people in the room? Three people? – Ethan Bolker Apr 22 '17 at 12:43

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Lets say there are n people at party.For a person, there can be 0 handshakes , 1 handshake, 2 handshakes.....$n-1$ handshakes. Clearly it is impossible to have a person having no handshakes and another person having all handshakes. Hence there must exist at least two person at party having same number of handshake (using peigon hole principle).

And for b) Think of that like this way, if i shake my hand with you then you shake hand with me.Thus handshakes occur in pairs.Now mark handshakes separately for person having even no. of handshakes and person having odd no. of handshakes .This gives you the required result.

Try this ,place value of n say upto 10 to check your answer.

Abhash Jha
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