Question

Toward the end of the principals' meeting for the drug free community campaign project, a total of 21 handshakes were exchanged. Assuming each principal shakes hands once with the other principals, how many principals were present?

Answer 1

1+2+3+4+5+6 = 21
therefore there were six principals

Answer 2

Lets assume there were x number of principles then,
Total number of handshakes(following the given conditions)= 1+2+3+...+(x-1)
21= (x-1)/2 [1+(x-1)]
42=(x-1)x
x^2-x-42=0
x^2-7x+6x-42=0
(x+6)(x-7)=0
either x+6=0 =>x=-6 which cannot be true
SO, x-7=0 =>x=7

Hence, there were 7 principles

Answer 3

Each of the x principals shakes hands with x-1 principals. but it takes 2 people to make a handshake. So

\[\large \begin{align} {{x(x-1)} \over 2} &= 21 \cr x^2 - x &= 42 \cr x^2 -x -42 &= 0 \cr (x-7)(x+6)=0 \end{align} \]

Since negative principals makes no sense, we choose 7 and reject the other root (-6).