Handshakes pass germs. Fist bumps are better. But a "Namaste!" is the best!
N people came into a party to have some fun. It is known that some couples attended the party, but the number of people who came by themselves is odd. Anyone can give a handshake to whomever they want- it is not compulsory for anyone. Assuming that either people went to the party by themselves or along with their spouse/girlfriend/boyfriend, what is the probability that the number of people who gave an odd number of handshakes is even?
Details and Assumptions:
- No one entered the party in groups that were larger than 2.
- Couples who entered can give a handshake to whomever they came with- it is also counted.
- No one gives more than one handshake to the same person once they enter the party.
- And no one can give a handshake to themselves. That would just be weird to look at.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
This goes for any set of things in the world that have shaken hands or have done a binary greeting or an exchange of anything- throughout history! That's really awesome. Invariance is so cool!
Let’s say that the i t h person shook hands n i times. Take the sum of n i over all i. This sum has to be even because it takes two to shake hands. So, the number of people who have given an odd number of handshakes is always even!