At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

See Solution: How many people in party?

lets say there are n persons

first person shakes hand with everyone else: n-1 times(n-1 persons)

second person shakes hand with everyone else(not with 1st as its already done): n-2 times

3rd person shakes hands with remaining persons: n-3

first person shakes hand with everyone else: n-1 times(n-1 persons)

second person shakes hand with everyone else(not with 1st as its already done): n-2 times

3rd person shakes hands with remaining persons: n-3

So total handshakes will be = (n-1) + (n-2) + (n-3) +…… 0;

= (n-1)*(n-1+1)/2 = (n-1)*n/2 = 66

= n^2 -n = 132

=(n-12)(n+11) = 0;

= n = 12 OR n =-11

-11 is ruled out so the **answer is 12 persons**.

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Arjun says

For Basic understanding, split the problem in small number first.

Lets assume there are 3 people. ABC. So from these three we get 6 combinations(AB,AC,BA,BC,CA,CB), 6 can be written as 3*2=6,means n(n-1).

but in the above combinations, there is a double counting(AB, BA) is same. so we have to divide by 2, which results in n(n-1)/2. this is also a combination formula ncr which n!/r(n-r)!, r is considered as 2 because only two ppl can shake.

The_Geek says

we can think this problem as a complete graph problem.

In a 4 vertices complete graph, no. of edges are n(n-1)/2, i.e. 6

Similarly here n(n-1)/2=66

so n(n-1)=132,

hence n=12.

shubham says

Ans is 12 by te formula of e=mc^2 lol:p

vinay says

12 people in the party

(n)c(2) = 66.