We will first get the number of 4 sets placed in a certain order,
and then we will "unorder" them:
To start with we will assume we are to place the set of 1 on the far left,
a set of 17 in the left middle, one set of 17 in the right middle, and
a set of 17 on the far right.
We can choose the far left set in 52C1 = 52 ways, leaving 51 cards.
We can choose the set in the middle left in C(51,17) ways, leaving 34 cards.
We can choose the set on the middle right in C(34,17) ways, leaving 17 cards.
Those 17 make up the set on the far right
So that would be 52*C(51,17)*C(34,17)*1
We're not done. There are 4! ways any group of four sets can be ordered,
so we must divide by 4! to unorder them.
Answer = 
Edwin
