The partitions should be used when they help solve the problem at hand. They have no physical significance and it is always possible to solve a problem without mentioning them. But sometimes they make it much easier to solve a problem. For instance, suppose you are calculating the number of ways that you can put two Fermions in three boxes, since each box can contain either zero or one fermion, than you can count them as follows
1: (1|1|0)
2: (1|0|1)
3: (0|1|1)
for a total of three states.
You can see that the first fermion can go in anyone of the three boxes, and second fermion can go on any of the remaining 2 boxes. 3*2=6= 3!/1!. At the ed you must divide by 2! because the fermions are identical so permutations among them won't create new states. You have then a total of 3!/(2!*1!)= 3 states. That can be generalized to
# of states = N!/[(N-n)!*n!] for n fermions on N boxes.
Now suppose you have two bosons that need to be placed in three boxes. since you can have more than one boson per boxe (no exclusion principle applies to bosons), there will be more states. Let's count them
1: (2|0|0)
2: (1|1|0)
3: (1|0|1)
4: (0|2|0)
5: (0|1|1)
6: (0|0|2)
for a total of six states. How can we generalize that? Doesn't seem obvious at first.
Here comes the math trick. Let's represent the particles by a '*'. the six states can now be represented as
1: (**||)
2: (*|*|)
3: (*||*)
4: (|**|)
5: (|*|*)
6: (||**)
Physically the '*' represents a particle while the '|' represents nothing whatsoever, it is just part of the notation. But that doesn't matter. Mathematically you have four symbols, Two '*'s and two '|'s. This symbols can be placed in 4!=24 different orders, but since the '*'s are identical we must divide by 2!, and the '|'s are also identical so we also divide by 2!. You have then a total of 4!/(2!*2!)= 6 states. That can be generalized to
# of states = (N+n-1)!/[(N-1)!*n!] for n bosons in N boxes.
Voila!