Pigeonhole Principle & Code Golf

Mathematica, 68 bytes

Print/@(10^RandomSample@RandomChoice[IntegerPartitions[+##,{#}]-1])&

An unnamed function that takes two integer arguments, the number of boxes, followed by the number of items.

It first computes all possible partitions of N+M into exactly M positive parts, and subtracts 1 from each partition afterwards. This gives us all possible partitions of N into M non-negative parts (which IntegerPartitions wouldn't generate otherwise). Then pick a random partition and shuffle it. This guarantees all possible ordered partitions with zeros are allowed. Finally, convert each bin of the partition to an output line by raising 10 to the corresponding power (such that each line becomes 1000... with k zeros). An example output might look like:

100
10000
1
10
10

Python 2, 77 86 bytes

from random import*
n,m=input()
exec'c=randint(0,n);n-=c;print 10**c;'*~-m
print 10**n

Generates a number in [0, n], prints that many items and substracts it from n. It does this m times.

This makes it very unlikely anything makes it to the last box, but the question only asked that every output be possible, not equally likely.

Batch, 164 bytes

@for /l %%i in (1,1,%1)do @set h%%i=1
@for /l %%j in (1,1,%2)do @call set/ab=%%random%%%%%%%1+1&call set/ah%%b%%*=10
@for /l %%i in (1,1,%1)do @call echo %%h%%i%%

I think 7 consecutive % signs might be a new personal best! Note: this produces odd output should it ever assign more than 9 items to the same box; if that's a problem, then for 180 bytes:

@for /l %%i in (1,1,%1)do @set h%%i=1
@for /l %%j in (1,1,%2)do @call set/ab=%%random%%%%%%%1+1&call call set h%%b%%=%%%%h%%b%%%%%%0
@for /l %%i in (1,1,%1)do @call echo %%h%%i%%

Yes, that's 28 %s in total on the second line.