Find the number of ways in which a list can be formed of the order of the 24 boats There are 15 rowing clubs;two of the clubs have each 3 boats on the river;5 others have each 2 and the remaining eight have each 1;find the number of ways in which a list can be formed of the order of the 24 boats,observing that the second boat of a club cannot be above the first and the third above the second.How many ways are there in which a boat of the club having single boat on the river is at the third place in the list formed above?

The number of ways in which a list can be formed of the order of the 24 boats $=24!$ but i cannot interpret what the second boat of a club cannot be above the first and the third above the second means.
The answers given is $\frac{24!}{(3!)^2(2!)^5},\binom{8}{1}\frac{23!}{(3!)^2(2!)^5}.$
 A: At first a boat is chosen from one of the $8$ clubs that only have a single boat, and this boat gets number $3$ on the list. This can be done in $8$ ways.
Then $23$ boats are left and if they are placed unconditionally that gives $23!$ possibilities. Looking at the conditions for a club that has $3$ boats we find only $1$ on the $3!=6$ orderings satisfies the condition. Now looking at the conditions for a club that has $2$ boats we find only $1$ on the $2!=2$ orderings satisfies the condition.
Repairing this gives a total of: $$8\times\frac{23!}{3!^22!^5}$$ orderings that satisfy all conditions.
A: HINT: You can interpret the question as the following:

In the image above, suppose we have a stack of piles and we can't take the pile $3$ without taking $1$ and $2$; and we can't take the pile $2$ without taking $1$ (because pile $3$ is under pile $1$ and $2$ and pile $2$ is under pile $1$). In this case, notice that there is only $1$ way of putting them to another rod (in order of $3-2-1$ from top to bottom).
In your question, the logic is the same. Suppose we have stacks of numbered piles (from $1$ to $24$) and piles are distributed so that there are two stacks with $3$ piles, five stacks with $2$ piles and eight stacks with $1$ pile. Then we are trying to put these $24$ piles into a single rod.
