boxes and stacking them options If I have 5 different boxes, lets give them a letter A, B, C, D, E and I have to stack them. How many different ways are there to stack them. If I have to use them all it would be 5!, however if I do not have to use them all for a stacking. So A is a solution, B is a solution, etc.
Now to add to this question, if a stacking of AB is the same as stacking as BA what would be the total number of stacking possible?
Hope this is clear. Hope you can help! Thank you :)
I know that for 2 boxes, A and B the result would be: A, B, AB (AB=BA)
I know that for 3 boxes, A, B and C the result would be: A, B, C, AB, AC, BC, ABC
Now I don't know how to continue.....
 A: It looks as if the order of stacking does not matter. So line up the boxes in order, and decide for each YES or NO, depending on whether we will use the box or not. There are $2^5$ ways to make the decisions. But one of the decisions we have just counted is the decision to use no boxes. That is presumably not allowed, so the answer to your problem is $2^5-1$.
If order does matter, then we can choose to make a stack of $1$ box, $2$ boxes, $3$ boxes, $4$ boxes, or $5$ boxes.
There are as you observed $5!$ ways to stack $5$ boxes. 
If we will use $4$ boxes, the boxes we use can be chosen in $\binom{5}{4}$ ways and then for each such choice they can be stacked in $4!$ ways, for a total of $\binom{5}{4}\cdot 4!$.
Similarly, for $3$ boxes there are $\binom{5}{3}\cdot 3!$ ways to do the job. And so on. Now add up.  
A: This smells like homework. Is it?
$5Cx$ gives the total number of combinations of 5 boxes taken $x$ at a time without caring for order. $$nCx = \dfrac{n!}{(n-x)!x!}$$
(Notice 5C2 = 10)
$5Px$ gives the total number of combinations of 5 boxes taken $x$ at a time but caring for order. $$nPx = \dfrac{n!}{(n-x)!}$$
(Notice 5P2 = 20)
In this case,
$$5P2 = 2 \times 5C2$$
Helps?
