Final answer:
To show there are c(n,2) handshakes among n people where each person shakes hands once with each other, we sum the sequence (n-1), (n-2), ..., 2, 1, which equates to n(n-1)/2, the formula for c(n,2) combinations.
Step-by-step explanation:
We need to show that in a group of n people, where each person shakes hands with every other person, there are c(n,2) different handshakes. The notation c(n,2) represents the combination formula C(n, k) = n! / (k! * (n-k)!), where n is the total number of items, and k is the number of items to choose.
When person 1 shakes hands with the other n-1 people, that's n-1 handshakes. Person 2 then shakes hands with everyone except for person 1 (who they've already shaken hands with), resulting in n-2 additional handshakes. This continues on with the third person shaking n-3 times and so forth, until the second to the last person shakes hands only once. The pattern of handshakes follows: n-1, n-2, ..., 2, 1. The total number of handshakes can be found by summing this sequence.
The sum of this arithmetic sequence is (n-1) + (n-2) + ... + 2 + 1, which equals to n(n-1)/2. This is exactly the formula for the combination of n items taken 2 at a time: c(n,2) = n(n-1)/2.
Therefore, for a group of n people, there are exactly c(n,2) different handshakes when each person shakes hands with others.