Given that there are two types of hats, red and blue, and the count of the red one is $m$ and the count of the blue one is $n$. Now, in how many ways can we arrange the hats in such a way that at any point in the queue, starting from the beginning, the number of red hats encountered is strictly more that the number of blue hats encountered ?
Given $n\geq2$ and $m\geq1$ and $m\gt n$.
Can anyone help me please. Thank you for help
The total number of arrangements is $$\frac{(m+n)!}{m!n!}.$$ As @lulu has given you, the probability that Red is always ahead is $$\frac{m-n}{m+n}$$ so just multiply these two expressions together.