12 man participate in work Recruitment Service. Each participant saw every other participant and shook his hand.

How many handshakes took place?


closed as off-topic by McMagister, xnor, BmyGuest, Tryth, mdc32 Jan 11 '15 at 23:07

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    $\begingroup$ I have edited your question to make the English clearer. Please check if it still matches your intent. $\endgroup$ – frodoskywalker Jan 10 '15 at 10:32
  • $\begingroup$ Please use appropriate tags. There is no need for a question to have a minimum(and maximum) of 5 tags. $\endgroup$ – Spikatrix Jan 10 '15 at 10:49
  • $\begingroup$ mathsisfun.com/combinatorics/combinations-permutations.html $\endgroup$ – A E Jan 10 '15 at 16:04
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    $\begingroup$ This is a standard combinatorics question that belongs on Mathematics SE instead. $\endgroup$ – xnor Jan 10 '15 at 18:52
  • $\begingroup$ @xnor he should still accept an answer. $\endgroup$ – warspyking Jan 11 '15 at 11:30

This is just math...


The numbers each represent a person. When we move on to person 2, he does not shake 11 times. He'll shake 10, due to already shaking with 1 earlier. This will repeat, which makes the equation super simple:


(0 because by the time it's 12's turn he's shaken with everyone already)
Which comes out to be:



66 handshakes will go down between those 12 men. 78 if people shake their own hands (lol)


This one is easy.

Each person needs to shake hands with 11 other participants and the total number of participants are 12. So the total number handshakes that took place would be

11x12=132 handshakes

But then there are some people who have shook their hands twice! So just divide the answer by 2 thus giving us

132/2=66 handshakes

and that is the answer. This is done because the first participant shakes their hand with 11 other participants,the second participant shakes their hand with 10 other participant(-1 because he has already shook hands with the first participant) and so on.. Thus the answer is

11+10+9+8+7+6+5+4+3+2+1+0=66 handshakes

The +0 in the answer is because the last participant has already shook his hands with all the other participant and so he does not have to shake hands with anyone.

  • $\begingroup$ Cool Guy in your method some people shake twice. If person A shakes with everyone and the person B does the same, A and B have shaken twice. See my answer. $\endgroup$ – warspyking Jan 10 '15 at 10:50
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    $\begingroup$ @warspyking , Your answer includes all the participants shaking his own hand.... $\endgroup$ – Spikatrix Jan 10 '15 at 11:08
  • $\begingroup$ Snap didn't even see that lol. $\endgroup$ – warspyking Jan 10 '15 at 11:09
  • $\begingroup$ Thank you, I've fixed my answer (Unless shaking their own hand is part of a trick or something.) $\endgroup$ – warspyking Jan 10 '15 at 11:10

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