At a party, everyone shook hands with everyone else, If there were 66 handshakes, how many people were at the party?

Let the number of persons= n

Number of handshakes by first person= n- 1

(As he cannot handshake with himself)

Number of handshakes by second person = n- 2

(As he has already handshaked with first person)

Number of handshakes by third person= n- 3

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Number of handshakes by second last i.e. ( n- 1)th person= n- (n- 1)= 1

Number of handshakes by last i.e. nth person= n- n= 0

So,

Total number of handshakes=(n- 1)+ (n- 2)+ (n- 3)+ 1+ 0

= 0+ 1+ 2+ 3+ ...........(n- 1)

= Sum of first (n- 1) terms

= (n- 1)(n- 1+ 1)/2

= n(n- 1)/ 2

So, Total number of handshakes= n(n- 1)/2

According to question,

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 or n= - 11(which is not possible as no. Of persons cannot be negat