10 Balls are to be placed randomly in 4 Boxes. What is the probability that any two boxes will contain exactly 2 and 3 balls? I am stuck at this question. I can't find that in how many ways we can place balls in boxes as given in the question (I think it has a mixed concept of Combination and Probability). I would like to know how to solve this type of question.
Out of 10, any random number of balls can be placed in one of the four boxes, then from the rest of the balls again a random number of balls are to be placed in the next box and repeat. Eg, there can be a, b, c and d number of balls in the four boxes such that a+b+c+d=10. The condition is that we choose a random box (say A) and then another box (let it be B). Then what is the possibility that box A and B will have exactly 2 and 3 balls respectively.
 A: The basic idea is the use of a multinomial expression.  The typical term is $\frac{n!}{i!j!k!m!}$ where $i+j+k+m=n$ and all indices are $\ge 0$.  The normalization to get probability is $\frac{1}{4^n}$.  In your case $n=10$ and you need to sum all those terms where one index $=2$ another index $=3$ and then divide by $4^{10}$.
A: I believe you can take this approach.  Consider the case of exactly 2 and exactly 3 balls out of 4 boxes filled by 10 balls at random.  First note that the underlying sample space has $4^{10}$ possible outcomes.  A way to think about this is the following:
(Case 1):
|Box1|  Box2 |  Box3  $\;\;\;\;\;$ | Box4 |
| _ _ | _ _ _ | _ _ _ _ _ |empty|   
(Case 2):
|Box1|  Box2 | Box3$\;\;\;\,$|Box4|
| _ _ | _ _ _ | _ _ _ _ | $\;\;$_$\;\;$|    
where the two above diagrams show the two basic possibilities given 10 balls and 4 boxes with exactly 2 in one and 3 in another.
Now we take the number of permutations in the first case:
Case (1) would be (if you consider each ball being associated with a 4-sided die, and the box that the ball is placed in the number showing on the die):
Ball#: 1 2 3 4 5 6 7 8 9 10
Box #: 1 1 2 2 2 3 3 3 3  3
What is the number of permutations of the above box pattern?  (the 1 1 2 2 2 etc.).
$10!/(2!3!5!) = 2520$
Then there are 4! patterns of (1):  $4!*2520= 60480$ total cases.
Case (2) would be like the below:
Ball#: 1 2 3 4 5 6 7 8 9 10
Box #: 1 1 2 2 2 3 3 3 3  4
The number of permutations is:
$10!/(2!3!4!1!)$ and again we have 4! patterns so a total of 302400 cases.
Combining result 1 and 2 yields: 362880.  Each of these has $(1/4)^{10}$ of occurring so the final probability of finding exactly one box with 2 and exactly one box with 3 balls is:
$362880*(1/4)^{10}=0.3460$
I hope this helps.
