Everybody in a room shakes hands with everybody else.

Question:

Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is

(a) 11

(b) 12

(c) 13

(d) 14

Solution:

Let us suppose there are n persons in a room for a handshake, 2 people are requried who will shake hand with each other.

So, number of ways to make a pair out of n persons is nC2 = 66 (given) 

i. $\mathrm{e} \frac{n !}{2 !(n-2) !}=66$

i. $\mathrm{e} \frac{n(n-1)(n-2) !}{2 \times(n-2) !}=66$

i.e $n^{2}-n=132$

i.e $n^{2}-n-132=0$

i. e $n^{2}-12 n+11 n-132=0$

i.e $n(n-12)+11(n-12)=0$

i.e $(n-12)(n+11)=0$

i.e $n=12$ or $n=-11$

Since n = −11 is not possible  

⇒  n = 12

Hence, the correct answer is option B.

 

 

 

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