What is the number of people who attended a party where every person shook hands with every other person if the total number of handshakes was 66?
Handshake Puzzle answer
Assuming that there are n people at a gathering and each person shakes hands with everyone else exactly once, the total number of handshakes can be calculated as follows:
- The first person shakes hands with n-1 people.
- The second person shakes hands with n-2 people (excluding the first person who they have already shaken hands with).
- The third person shakes hands with n-3 people (excluding the first and second persons who they have already shaken hands with).
- This pattern continues until the last person shakes hands with one person (excluding everyone else).
Thus, the total number of handshakes is the sum of the first n-1 positive integers, which can be expressed as (n-1)+(n-2)+(n-3)+…+3+2+1.
Equating this expression to the given value of 66 gives the equation:
n(n-1)/2 = 66
Solving for n yields:
n = 12
Therefore, there were 12 people at the gathering.
Another question based on this
The Handshake Puzzle is a classic problem in combinatorics that involves calculating the total number of handshakes that occur when a group of people shakes hands with each other. The problem assumes that each person in the group shakes hands exactly once with every other person in the group.
To solve this problem, you can use a simple formula that relates the number of handshakes to the number of people in the group. The formula is:
(number of handshakes) = (number of people) x (number of people – 1) / 2
For example, if there are 4 people in a group, the total number of handshakes would be:
(4 x 3) / 2 = 6
This means that there would be a total of 6 handshakes in a group of 4 people.
If there are n people in a group, the formula can be generalized as:
(number of handshakes) = n(n-1)/2
So, if there are 10 people in a group, the total number of handshakes would be:
10(10-1)/2 = 45
Therefore, there would be a total of 45 handshakes in a group of 10 people.