Handshakes again?!

I completed this problem using Mathematical Induction.

The formula for this combination is = $\frac{n}{2} (n-3)$

STEPS TO DERIVE FORMULA

1) Draw out the smaller combinations on paper. First find how many handshakes there are for a table with 4 people. Then do the same with a table of 5 people ... and so on...

2) As you draw the tables, a patterns starts to appear. If the people were to take turns shaking hands - The the first two people always shakes hands with an "extra" $(n-3)$ people. While each successive person shakes hands with an "extra" of the $(previous - 1)$ . This continues and the last 2 people shakes with an extra 0 people.

e.g. The pattern for a table with 6 people -

• 1st person shakes with - 3 extra
• 2nd person shakes with - 3 extra
• 3rd person shakes with - 2 extra
• 4th person shakes with - 1 extra
• 5th person shakes with - 0 extra
• 6th person shakes with - 0 extra

Note: Here (n-3) = (6-3) = 3;

3) Derive a Series - For a table with "n" people, the number of shakes would be -

$(n-3) + \underline{(n-3) + (n-4) + (n-5) ... + (n-n)} + (n-n)$

"Excluding" the first element and last element, the series is: "The Sum of All Integers upto n"

4) Derive formula - The sum of integers upto $n = \frac{n}{2}(n+1)$ Therefore, the sum of ALL the handshakes =

$(n-3) + \frac{n}{2}(n+1) + 0$

5) Use algebra to simplify or standardize the formula. And voila - you've got the answer.