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Question: A total of 28 handshakes were exchanged at the conclusion of a party. Assuming that each participant...

A total of 28 handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite towards all the others, the number of people present was:

Explanation

Solution

In this question it is given that the total number of handshakes is 28. So, if we consider that there are ‘n’ number of people. Then the first person shakes hands with (n-1) persons (he can’t shake hands with himself so we have subtracted 1 that n-1). Similarly, the second person shakes hands with (n-2) persons (n-2 because the second person can’t shake hands with himself and the first person is already counted), and so on. At last apply formula, total number of handshakes(n−1) +(n−2) +…. +1=n(n−1)2\left( {n - 1} \right){\text{ }} + \left( {n - 2} \right){\text{ }} + \ldots .{\text{ }} + 1 = \dfrac{{n(n - 1)}}{2}.

Complete step-by-step answer: Given,
Total handshakes were exchanged at the conclusion of a party = 28.
We have, total number of handshakes = (n−1) +(n−2) +…. +1=n(n−1)2\left( {n - 1} \right){\text{ }} + \left( {n - 2} \right){\text{ }} + \ldots .{\text{ }} + 1 = \dfrac{{n(n - 1)}}{2}
Using the above formula, we get;
n(n−1)2=28\dfrac{{n(n - 1)}}{2} = 28
⇒n(n−1)=56\Rightarrow n(n - 1) = 56
⇒n2−n−56=0\Rightarrow {n^2} - n - 56 = 0
⇒n2−8n+7n−56=0\Rightarrow {n^2} - 8n + 7n - 56 = 0
⇒n(n−8)+7(n−8)=0\Rightarrow n(n - 8) + 7(n - 8) = 0
⇒(n−8)(n+7)=0\Rightarrow (n - 8)(n + 7) = 0
⇒n=8\Rightarrow n = 8 or ⇒n=−7 \Rightarrow n = - 7
Negative number of people is not possible. So, n = -7 is not acceptable.
Therefore, n = 8 is the only acceptable value.
So, the number of people present in the party = 8.

Note: In the above question first person shakes hand with 7 persons, second person shakes hand with 6 persons; similarly, third person shakes hand with 5 persons and so on.
Total handshakes = 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 Handshakes.
There is an alternative method which can be used in this question by using CombinationnC2=n!2!(n−2)!=n(n−1)2^n{C_2} = \dfrac{{n!}}{{2!(n - 2)!}} = \dfrac{{n(n - 1)}}{2}.
Above equation has a numerical value of 28.
Thus, we get the equation;
n(n−1)2=28\dfrac{{n(n - 1)}}{2} = 28
Solving this equation, we get two values, choose the appropriate value.
Subsequent steps are already covered in the main answer. This is the alternative method. We can follow any of the steps for solving this question.