Question

Everybody in the room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is?
A) 11
B) 12
C) 13
D) 14

Answer 1

Hint: In the question mentioned above, we need to recall the concept of permutations and combinations. Permutation in mathematics is defined as an act in which we arrange all the members of the set into some type of sequence or order. Combination in mathematics is defined as the process of selecting certain items to form a collection. Unlike permutation, the combination does not have to be in any specific order.

Formula used:
The formula to calculate permutation is given as:
\[{}^n{P_r} = \dfrac{{(n!)}}{{(n - r)!}}\]
Where, permutation occurs from the choice of the given \[r\] things from the \[n\] set of things.
The formula to calculate the combination is as follows:
\[{}^n{C_r} = \left(\dfrac{n}{r}\right) = \dfrac{{{}^n{P_r}}}{{r!}} = \dfrac{{n!}}{{r!(n - r)!}}\]
In which, \[r\] is usually the choice and \[n\] things represent the set.

Complete step-by-step solution:
Let us assume that the total number of people in the room is \[n\]
To make a single handshake we need two people, so
The number of handshakes would be \[ = {}^n{C_2}\]
So, \[{}^n{C_2} = 66\]
\[\dfrac{{n!}}{{2!(n - 2)!}} = 66\]
\[\dfrac{{n(n - 1)(n - 2)!}}{{2 \times 1 \times (n - 2)!}} = 66\]
\[\dfrac{{n(n - 1)}}{2} = 66\]
\[{n^2} - n = 132\]
\[{n^2} - n - 132 = 0\]
\[{n^2} - 12n + 11n - 132 = 0\]
\[n(n - 12) + 11(n - 12) = 0\]
\[(n - 12)(n + 11) = 0\]
\[n - 12 = 0,\,n + 11 = 0\]
\[n = 12\,,\,n = - 11\]
Therefore \[n = 12\], since \[n\] cannot be equal to \[ - 11\].
Hence, option B , 12 is the correct answer

Note: Permutation is used where the order of the data that is collected does matter, whereas combination is used where the order of the data collected does not matter. Permutation can be used for the arrangement of people, numbers, letters colours, etc, whereas combination can be used for the food menu, clothes, subjects, team, etc where no particular order is required.