# Confused on what to do here..

• January 23rd 2008, 11:59 AM
eh501
Confused on what to do here..
Five friends met for lunch, and they all shook hands. Each person shook the other persons right hand only once. What was the total number of handshakes?

• January 23rd 2008, 12:32 PM
ThePerfectHacker
Quote:

Originally Posted by eh501
Five friends met for lunch, and they all shook hands. Each person shook the other persons right hand only once. What was the total number of handshakes?

No. Let A B C D E be the people. Then A shakes hands with B,C,D,E. Then B shakes hands with C,D,E but not A because they already shook hands. Then C shakes hands with D,E. And D shakes hands with only E.
• January 23rd 2008, 12:34 PM
eh501
How would you know when to do that, when to multiply or when to add?
• January 23rd 2008, 12:37 PM
ThePerfectHacker
Quote:

Originally Posted by eh501
How would you know when to do that, when to multiply or when to add?

Each problem is different, you need to understand how to approach a problem and decide whether you have to add or multiply.

You can also do it another way. Draw a picture and the lines represent the people who shook hands. Then count the number of lines.
• January 23rd 2008, 12:42 PM
eh501
• January 23rd 2008, 12:52 PM
ThePerfectHacker
Quote:

Originally Posted by eh501

You are going to find the answer. I basically gave you the solution.
• January 23rd 2008, 12:55 PM
eh501
I know you did but i really don't know what to do..do i just add up the letters? I still came up with 10.
• January 23rd 2008, 12:58 PM
ThePerfectHacker
Quote:

Originally Posted by eh501
I know you did but i really don't know what to do..do i just add up the letters? I still came up with 10.

Yes, 10 is the right answer. If you add up the lines. But your reasoning that 5+5=10 is wrong. Try for example 4 people instead of 5. Then the answer would be 3+2+1 = 6 and not 4 + 4 = 8.
• January 23rd 2008, 02:34 PM
JaneBennet
If n people shake hands with each other such each pair of persons shake hands once and only once, the total number of handshakes is $^n\mathrm{C}_2=\frac{n(n-1)}{2}$.