Handshakes

[katolyn]

I have a similar problem to a previous one about handshakes at a reunion, however, I am somewhat mathematically challenged and really need someone to spell out how to set the problem up. I have been grappling with this for over an hour now. Thanks for your help!

Problem: There are 25 people in the room. To get acquainted, each person shakes hands with each of the other people. each pair of people shakes hands once and only once. what is the total number of handshakes?

 

[wjm11]

Problem: There are 25 people in the room. To get acquainted, each person shakes hands with each of the other people. each pair of people shakes hands once and only once. what is the total number of handshakes?

Just google something like "math handshake". One example:

"A good way to model the situation is to think of an n-sided polygon,
which has n vertices.

Now consider how many diagonals you can draw between the vertices, and
also include two lines connecting a particular vertex to the two
adjacent ones. It is clear that there are (n-1) lines joining any one
vertex to the other vertices in the polygon. If we consider each of
the n vertices, each requires (n-1) lines to join to the other
vertices. There are thus n(n-1) links. But each of these links has
been produced twice, once from each end, and so the number n(n-1) is
too large by a factor of 2.

Hence total number of connections joining every vertex to every other
one is:

n(n-1)
------
2

http://mathforum.org/library/drmath/view/56157.html

 

[soroban]

Hello, katolyn!

There are 25 people in the room.
To get acquainted, each person shakes hands with each of the other people.
Each pair of people shakes hands once and only once.
What is the total number of handshakes?

We can baby-talk our way through this . . .

Let's say there are 25 men: .\(\displaystyle A,B,C,D,\,\hdots\,Y\)


Select any man, say, \(\displaystyle A.\)
He will shake hands with each of the other 24 men.
So there are 24 handshakes: .\(\displaystyle AB,\:AC\:AD,\:\hdots\:AY\)

Select another man, say, \(\displaystyle B.\)
He will shake hands with each of the other 24 men.
So there are 24 handshakes: .\(\displaystyle BA,\:BC,\:BD,\:\hdots\:BY\)

Selet yet another man, say, \(\displaystyle C.\)
He will shake hands with each of the other 24 men.
So there are 24 handshakes: .\(\displaystyle CA,\:CB,\:CD,\:\hdots\:CY\)

Then for the 25 men, there will be: .\(\displaystyle 25\times 24 \:=\:600\) handshakes.


But our list contains \(\displaystyle AB\) and \(\displaystyle BA\), which is the same handshake.
. . ("\(\displaystyle A\) shakes hands with \(\displaystyle B\)" is the same as "\(\displaystyle B\) shakes hands with \(\displaystyle A\).")
And the list contains \(\displaystyle \{AC,\,CA\}\) and \(\displaystyle \{BC,\,CB\}.\)

Our list contains every handshake and its redundant reversal.
. . Hence, our answer is twice as large as it should be.

\(\displaystyle \text{The correct answer is: }\:\frac{25\cdot24}{2} \:=\:300\text{ handshakes.}\)


Of course, it is much faster to use the formula explained by wjm11.

. . \(\displaystyle \text{For }n\text{ people, there will be: }\:\frac{n(n-1)}{2}\,\text{ handshakes.}\)