In a queue of 4 men, 6 women and 20 kids, what is the probability that all men appear before the 2nd woman?
What will be a good way to approach this problem?
My (wrong) attempt using combinatorics
Count number of ways all 10 adults(A) can be arranged first into 2 groups (left vs right group)
Left group: must contain all 4 men and 1 woman
Right group: contains the remaining 5 women
The 20kids(K) can then be distributed among the 2 groups. Denote (left group, right group)
Below are the different ways K can be chosen and distributed, with permutation in each group accounted for.
Case1: (20K & 5A, 0K & 5A) = 20C20 x 25! x 5!
Case2: (19K & 5A, 1K & 5A) = 20C19 x 24! x 6!
Case3: (18K & 5A, 2K & 5A) = 20C18 x 23! x 7!
…
Then sum up all 20cases and multiply by 6, since the woman in left group could have been any of the 6 women. However, there's definitely overcounting – eg. case1 and case2 both counts the arrangement where all 20K are in the center (queue looks like this: 5A 20K 5A)
Best Answer
The positions of the children are irrelevant. What matters here are the relative positions of the men and women.
In any arrangement, of the $10$ positions occupied by the four men and six women, men must occupy four of the first five positions in order for all four men to appear before the second woman. Hence, the probability that all four men appear before the second woman is $$\frac{\dbinom{5}{4}}{\dbinom{10}{4}}$$