Question

Everybody in a room shakes hands with everybody else. The total number of handshakes is 45. The total number of persons in the room is
[CET - 2018]​

Answer 1

Let the people in a room be 'n'

For the full answer refer the attachment..

The answer is 10 people.

Answer 2

$\large\underline{\sf{Solution-}}$

We know that, when two persons shake hands with each other, it is counted as one hand shake.

Given that,

• There are 45 hand shakes total in a room.

Let suppose that number of persons in a room be 'n'.

So,

Total number of hand shakes is same as the number of ways of selecting 2 persons from 'n' persons and this can be done in C( n, 2 ) ways.

According to statement.

Total number of hand shakes = 45.

$\rm :\longmapsto\:C(n, 2) \: = \: ^nC_2 \: = \: 45$

We know,

$\boxed{ \sf{ \: ^nC_r \: = \: \dfrac{n!}{r! \: (n - r)!}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{n!}{2!(n - 2)!} = 45$

$\rm :\longmapsto\:\dfrac{n(n - 1) \: \cancel{(n - 2)!}}{2 \times 1 \: \cancel{(n - 2)!}} = 45$

$\rm :\longmapsto\:\dfrac{n(n - 1)}{2} = 45$

$\rm :\longmapsto\:n(n - 1) = 90$

$\rm :\longmapsto\:n(n - 1) = 10 \times 9$

$\rm :\longmapsto\:n(n - 1) = 10 \times (10 - 1)$

So, on comparing we get,

$\bf\implies \:n = 10$

Hence,

$\: \: \: \: \: \: \: \: \purple{ \boxed{ \sf \: Number \: of \: persons \: in \: room \: is \: 10}}$

Additional Information :-

$\rm :\longmapsto\: ^nC_1 = \: ^nC_{n - 1} = n$

$\rm :\longmapsto\: ^nC_0 = \: ^nC_{n} = 1$

$\rm :\longmapsto\: ^nC_x = \: ^nC_{y} \implies x = y \: \: or \: n = x + y$

$\rm :\longmapsto\:^nC_r = \dfrac{n}{r} \: ^{n - 1}C_{r - 1}$

$\rm :\longmapsto\:^nC_r \: + \: ^nC_{r - 1} \: = \: ^{n + 1}C_r$

$\rm :\longmapsto\:\dfrac{^nC_r}{^nC_{r - 1}} \: = \: \dfrac{n - r + 1}{r}$