BOX Problem A combinatorics question A big box contains 10 small boxes. Each of these small
boxes is either empty or contains 10 boxes which are even
smaller. Again, each of these smaller boxes is either empty
or contains 10 boxes smaller than itself and so on. After
opening all the boxes it turned out that exactly 2008 boxes
were nonempty. So, how many boxes were empty ?
This a math competition question.  I solved it and think that it is 2007.  Could anyone confirm that
 A: There’s no reason to introduce probability, and I see no reason for the assumption that a box is non-empty with probability $\frac12$. Suppose that there are $e$ empty boxes and $n$ non-empty boxes. Imagine that each box except the big one is connected by a string to the box that contains it. There are $e+n$ boxes altogether, but one of them is the big one, so there are $e+n-1$ strings. On the other hand each non-empty box has $10$ strings attaching it to the smaller boxes inside it, and this accounts for all of the strings, so there are $10n$ strings. Thus, $e+n-1=10n$, so $e=9n+1$. In our case $n=2008$, so $e=18073$.
This is really a problem in graph theory. The boxes form a rooted tree, the big box being the root. The non-empty boxes correspond to interior nodes of the tree and the empty boxes to leaves. Each interior node has $10$ daughter nodes. The strings are the edges of the tree.
A: Probability of a box being empty = 1/2
Probability of a box being non-empty = 1/2
Level 1 Let n1 be the number of  boxes empty, then 10-n1 will be the numbe non-empty
P[x=n1] = 10Cn1 (1/2)^10
P[x=10-n1] = 10C(10-n1) (1/2)^10
Level 2 For each box that is non-empty, let there be n2 boxes empty and 10-n2 boxes non empty.
P[x=n2] = 10Cn2 (1/2)^10
P[x=10-n2] = 10C(10-n2) (1/2)^10
Recursively,
A => No of boxes empty = P[x=n1]P[x=n2]...(10*100*1000...
B => No of boxes non-empty = P[x=(10-n1)]P[x=(10-n2)] ... *(10*100*1000...
Using the property => nCr = nCn-r
A = B
In which case, 2008 were non-empty including the root box.  If we opened that will be the levels 1 , 2 , 3....  Thus the number of empty will be 2008-1 = 2007 ( deducting the root).
Do you think the approach is right?
Thanks
Satish
