Four politicians and four lawyers attend a party. Each politician shakes hands exactly once with everyone, and each lawyer shakes hands exactly once with each politician. How many handshakes take place?
HI! Wolf here.
Ok so, I will give you hints and also my solution.
So there are multiply ways that you can do it, most commly the algebraic way.
However I am not well versed in that method SO I will help u with a way that I learned how to to these questions!
1. Draw two rows with 4 dots in each
2. Every dot on the left is a politician and every dot on the right is a lawyer
3. Now draw a line one person in the lawyer collom to the other 4
4. Because the other people (politician) doesnt shake hands with the others, there is no overlap (I think)
So count those up! But the wording of the question is confusing me a bit. I am gonna assume that both of the handshake rules are in place at the same time. So that the politicians are still only shaking the hands once and that the lawyers also only shake there hand once too.
This means that, the only thing we have to figure out is the politicians.
The way we do this is by counting how many hands the first person shakes then how man the second one shakes.
HOWEVER!!!
Because the first person already shook the second person hand, the second person can NOT shake person 1 agian, making 1 shaking 3 hands!! (also beliving that they dont shake there own hand)
Same with 3, the cant sake 1&2. and 4 cant shake anyone.
Add the both numbers together and you will get your answer!
I cant figure out a way to hide my answer to here it is please look over the actuall way to solve it...
16+6= 22
I hope this helps!!!
~Wolf
