TITLE: Probability - Brick in box
QUESTION [7 upvotes]: From the set {1, 2, 3, ... 999}, 6 distinct numbers are chosen. These are divided into two groups $a_1,a_2,a_3$ and $b_1,b_2,b_3$. Find the probability that a brick made from the dimensions of group $a$ fits into a box made from the dimensions of $b$. Assume that the brick can be rotated in a suitable manner to be made to fit inside the box

I am able to comprehend this question and what it asks. However, I don't have a strategy in place to solve this question. The solution provided has the following to say:

without loss of generality, we can say that out of the 20 possible cases, 5 are suitable. Therefore, the probability is $\frac 1 4$.

What are the twenty possible cases here? How can we set conditions on $a$ and $b$ such that we can obtain these cases? The ability to rotate the brick before placing in the box has confused me. Any help would be appreciated!

REPLY [7 votes]: The $20$ cases are given by $C^{6}_{3} = 20$. They are ways of splinting a set of $6$ distinct numbers into $2$ sets of $3$ numbers. 
Ie. consider  $\{a_{1},a_{2},a_{3},a_{4}, a_{5},a_{6}\}$ to be in increasing order. Then the $5$ good cases are:  
$(\{a_{1},a_{2},a_{3}\},\{a_{4}, a_{5},a_{6}\})$
$(\{a_{1},a_{2},a_{4}\},\{a_{3}, a_{5},a_{6}\})$
$(\{a_{1},a_{2},a_{5}\},\{a_{3}, a_{4},a_{6}\})$
$(\{a_{1},a_{3},a_{4}\},\{a_{2}, a_{5},a_{6}\})$
$(\{a_{1},a_{3},a_{5}\},\{a_{2}, a_{4},a_{6}\})$
Edit: there should be ways of proving there are only $5$ without checking them all. Here is one method: $a_1$ must be part of the first set because if it were part of the second set, there is no number smaller than $a_1$, so the brick wouldn't fit. Similarly, $a_6$ must be part of the second set. It remains to split the remaining $4$ numbers $\{a_{2},a_{3},a_{4},a_{5}\}$ into $2$ sets of $2$. We have $C^{4}_{2} = 6$ ways of doing this. Five of those give us the above sets. The sixth one is $(\{a_{4},a_{5}\},\{a_{2},a_{3}\})$ giving  $(\{a_{1},a_{4},a_{5}\},\{a_{2},a_{3},a_{6}\})$ 
which does not describe fitting bricks because $a_{4}>a_{3}$.