How to Solve the Handshake Counting Technique?

Question

Its 2 problem separated.

Suppose you are organizing a business meeting and are in charge of facilitating the introductions.

 

A. Suppose there are 5 people in the group
1.How would you arrange the group  so each person can shake hands with every other person?
2. How many times will each person shake hands with someone else?
3. How many handshakes will occur?

 

B. Suppose there is an unknown number (n) people in the group

1.How would you arrange the group of n people so each person can shake hands with every other person?
2. How many times will each person shake hands with someone else?
3. How many handshakes will occur? This answer should be expressed as a general counting formula that depends on the value of n.

Answer 1

Let's say the 5 people are labelled A, B, C, D, E. 

PART A

1) A shakes hands with B, C, D, E then moves away from the group (4 handshakes)

B shakes hands with C, D and E then moves away from the group (3 handshakes)

C shakes hands with D and E then moves away from the group (2 handshakes)

D shakes hands with E (1 handshake)

3) 10 handshakes in all. Everyone has shaken hands with everyone else.

2) Example: take C: C shook hands with A when before A moved away from the group; C shook hands with B before B moved away from the group. C proceeds to shake hands with D and E before leaving the group. This example shows that each person shakes hands with 4 others.

PART B

1) Assemble participants in arbitrary order. The first in order shakes hands with n-1 others then leaves the group.

The next in order shakes hands with n-2 others and leaves the group.

And so on until only two remain to shake hands. The exercise has then been completed.

2) Each participant shakes hands with n-1 others. When n=5 we get 4 handshakes,  which ties in with Part A.

3) The summation of handshakes is n-1+n-2+n-3+...+2+1, which can be written (n-1+1)+(n-2+2)+(n-3+3)+... There are (n-1)/2 of these pairings so the summation is n(n-1)/2. If n=5 we get 5*4/2=10 handshakes which ties in with Part A. If n=6 we have 5+4+3+2+1=15=(6-1)*6/2=30/2. The counting formula is n(n-1)/2.