Note that each term looks like x^a y^b z^c where a,b, and c are nonzero integers. Also, since they come from (x+y+z)^100, we have the equation a+b+c=100.
The easiest way I know to count the terms is the "stars and bars" counting method.C(n+r-1,n)=C(n+r-1,k-1)=C(100+3-1,2)
Consider a line of 102 boxes. Choose two of the boxes. Set a = the number of boxes to the before the first box chosen, b = the number of boxes between the two boxes chosen, and c = the number of boxes after the second box chosen.
It can then be shown that this situation is in bijection with the terms in the expansion of (x+y+z)^100.
Therefore, there are 102C2 = 5151 terms in the expansion of (x+y+z)^100