If $B_{i}$ denotes number of balls in $i^{th}$ box then find the probability that $B_{1}<B_{3}<B_{5}<B_{7}$ and $B_{2}>B_{4}>B_{6}>B_{8}$.

Question

$36$ identical balls are randomly distributed in $8$ different boxes in such a manner that no box is empty and no two boxes have same number of balls.

If $B_{i}$ denotes number of balls in $i^{th}$ box then find the probability that $B_{1}<B_{3}<B_{5}<B_{7}$ and $B_{2}>B_{4}>B_{6}>B_{8}$.

My Attempt I am facing difficulty in finding number of ways to distribute balls so that boxes contain unequal number of balls.Once this is done then rest shouldn't be very difficult.

Answer 1

In case you haven't noticed, note your conditions on the number of balls in each box means that, for the minimum number of balls required, there is a ball in one box, 2 balls in another box, and so on until you get 8 balls in the 8th box. If you add up this minimum set of balls and compare it to your number, i.e., 36, I think you will find something interesting, and potentially useful for solving the problem.

Answer 2

Much easier is to use the symmetry of the problem. For each arrangement of balls with $B_1 \lt B_3$ there is a corresponding one with all the rest of the balls in the same positions but $B_1 \gt B_3$ and vice versa. That tells you that with the conditions of your problem, the chance $B_1 \lt B_3$ is $\frac 12$. Extend this to a chain of four inequalities.