THIS IS FORMATTED

B is (2^m-1)^n

Firstly since all the presents need to have at least one "home", let's find and list those first :D

Consider In the case of n=2 presents and m=3 boxes: (X, Y are presents)
Also without loss of generality, we mark these as the FIRST OCCURRENCE of the present type
This is to avoid duplicate counting (X Y+X <-> X X+Y)
XY _ _
X Y _
X _ Y
Y X _
_ XY _
_ X Y
Y _ X
_ Y X
_ _ XY

Now in each of these case we think about the remaining ways to put the other presents
An example would definitely help explaining

Let's take "X Y _"
Since each present type is independent from each other we can consider them separately and multiply the number of ways together
Remember, X and Y marks the FIRST occurrence of the present type
X: In the last two boxes, each of them have two choices: Either you put a X, or you don't. That's 2^2 ways (X _ _, X X _, X _ X, X X X)
Y: In the last box, it also has two choices: You put a Y, or you don't
That's 2^1 way
So "X Y _" (as the "base") will create 2^2 x 2^1 cases as follow
X Y _ + _ _ _  = X  Y __
X Y _ + _ X _  = X XY __
X Y _ + _ _ X  = X  Y X
X Y _ + _ X X  = X XY X
X Y _ + _ _ Y  = X  Y  Y
X Y _ + _ X Y  = X XY  Y
X Y _ + _ _ XY = X  Y XY
X Y _ + _ X XY = X XY XY

Now notice that counting similarly,
BASE      2^X   2^Y
XY _ _ -> 2^2 x 2^2 ways
X Y _  -> 2^2 x 2^1 ways
X _ Y  -> 2^2 x 2^0 ways
Y X _  -> 2^1 x 2^2 ways
_ XY _ -> 2^1 x 2^1 ways
_ X Y  -> 2^1 x 2^0 ways
Y _ X  -> 2^0 x 2^2 ways
_ Y X  -> 2^0 x 2^1 ways
_ _ XY -> 2^0 x 2^0 ways
+)
------------------------
2^2x(2^2+2^1+2^0)+2^1x(2^2+2^1+2^0)+2^0x(2^2+2^1+2^0)=(2^2+2^1+2^0)^2
(Make sure you understand the example)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Finally you can generalise it a bit and think a bit to get
WAYS = (2^(m-1)+2^(m-2)+...+2^1+2^0)^n=(2^m-1)^n
Use some binary modulo exponent thing OR use python :)

Please read omg