Answer Posted / auswin

An alternate mathematical approach:
If 'n' people shake hand with each other, the first
person can shake hands with 'n-1' people, the second with
'n-2' people and so on till the last person with 'n-n i.e 0'
people.

So, the total number of handshakes are
n-1+n-2+n-3+....n-n which is nothing but n*n-(1+2+3+....n)
which is equal to
n^2 -(n(n+1)/2) = (2n^2-n^2-n)/2 = (n^2-n)/2 = n(n-1)/2 ..!!

Hence, the number of handshakes possible with n people is
n(n-1)/2......

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