Will not do the full induction here. Fill in the gaps
yourself.
Let P_k be the statement "the number of handshakes for n ppl in a
party where each person shakes another person's hand only once =
n(n - 1)/2" where n is in the set of natural numbers and n >
1
P(2) is trivial.
Assume P(k) is true for some k in the set of natural numbers and
again k > 1.
Now let us consider the case for k + 1 ppl. We consider the number
of hand shakes the k + 1 person has to make with the rest of the k
ppl in the grp. Clearly, he needs to shake k times.
Hence the number of total number hand shakes that occur = k(k -
1)/2 + k
= k(k + 1)/2
This implies P(k+1) is true.
Hence P(k) is true => P(k + 1)
P(2) => P(3) => P(4)...
Hence P(n) is true by M.I.
Quite trivial question. Try proving using combinatorics. The answer
is sweeter.
