Solution:
With two people (A and B), there is one handshake
(A with B).
With three people (A, B, and C), there are three handshakes
(A with B and C; B with C).
With four people (A, B, C, and D), there are six handshakes
(A with B, C, and D; B with C and D; C with D).
In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
n^2-n =132 n^2-n-132=0
(n-12)(n+11)=0
n =12 or n= -11
As handshakes cannot be negative we discard 11 . Therefore answer is , 12 people.
Each handshake involved 2 people so 12 guests: Guest A made 11 handshakes, B 10 having already been counted as an A-B handshake, C 9.
With two people (A and B), there is one handshake
(A with B).
With three people (A, B, and C), there are three handshakes
(A with B and C; B with C).
With four people (A, B, C, and D), there are six handshakes
(A with B, C, and D; B with C and D; C with D).
In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
n^2-n =132 n^2-n-132=0
(n-12)(n+11)=0
n =12 or n= -11
As handshakes cannot be negative we discard 11 . Therefore answer is , 12 people.
Each handshake involved 2 people so 12 guests: Guest A made 11 handshakes, B 10 having already been counted as an A-B handshake, C 9.