(28 votes, average: 3.61 out of 5)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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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.
How did you get the 4 vertices
how n(n-1) = 132 , results in 12
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.
Ans is 12 by te formula of e=mc^2 lol:p
12 people in the party
(n)c(2) = 66.