solve : 

For (A) with nn people at the party, what are the possible number of hands each person could shake?

Show that it is impossible for both someone to have shaken hands with everyone and someone to have shaken hands with noone simultaneously.

Is there some way to use the pigeon-hole principle here?

For (B) count the number of handshakes that occur by adding up how many handshakes people participated in individually and recognize that this overcounted somehow (because each handshake we counted twice: once from the shorter person of the two and again from the taller person). This gives us a useful result called "the handshaking lemma."

What happens then if there are an odd number of people who shook an odd number of hands? Remember that the number of handshakes must be a whole number.

In a party people shake hands with one another (not necessarily everyone with everyone else).
(a) Show that 2 people shake hands the same no. of times.
(b) Show that the number of people who shake hands an odd no. of time is even.

6x6=64

The best way I can think of is to seperate these rectangles into three groups:

a) 1x1 squares

b) 1xN rectangles, 1<N<=8

c) Nx1 rectangles, this should be the exact same number as in b)

d) MxN rectangles, 1<N,M<=8

The reason for this split is that i thought of a nice way to get group d):

I believe that every d) rectangle is uniquely defined by a pair of opposite corner squares.

The amount of possible such pairs is 64*49/2=1568,

with the reasoning of picking the first square, and then having 49 squares left to pick which are not on the same rank/file.

Lastly, each rectangle in d) has two possible diagonal pairs which uniquely define it.

Therefore, we have 1568/2=784 rectangles 8x8