I found a permutation problem like "show that there is a ... number of ways of..." which is given below. I tried to show that, but I got a different answer. I will describe how I did, after stating the problem.
Here is the problem:
There are $5$ roads connecting the cities $A$ and $B$. $3$ men want to travel from $A$ to $B$, then again $B$ to $A$. If the journeys $AB$ and $BA$ must be not on the same road and no two men can use the same road when traveling from one city to another city, show that there are $1440$ ways they can travel.
First, I found there are $5\times4\times3=60$ ways they can travel from $A$ to $B$. Similarly, without any special case, there are $60$ ways of traveling from $B$ to $A$. Considering one permutation of traveling from $A$ to $B$, there are $12+9+7=28$ different ways to travel from $B$ to $A$, when at least one man is using the same way. Because the number of different ways to return when
- the first man reuses a road is $4\times3=12$,
- the second man reuses but the first man does not is $12-3=9$,
- the third man reuses but the first and second do not is $12-3-2=7$.
Therefore, the total number of ways they can travel is $60\times(60-28)=1920$, which is not the result given.
What was the mistake? Please, explain the correct method. Sorry for the faults in explaining.
Thank you.
As Daniel Mathias stated in the comments, your solution is correct.
Here is another method of solving the problem:
As you observed, there are $5 \cdot 4 \cdot 3$ ways for the three people to take three different routes from $A$ to $B$.
Since there are six one-way trips and oly five available roads, at least one person must travel from $B$ to $A$ by a route another person took to get from $A$ to $B$.
Exactly one person travels from $B$ to $A$ by a route another person took to get from $A$ to $B$: There are three ways to select the person who travels from $B$ to $A$ by a route another person took to travel from $A$ from $B$, two ways for that person to select one of those two routes, and $2!$ ways for the remaining two people to travel from $B$ to $A$ by the two routes nobody took from $A$ to $B$. Hence, there are $$3 \cdot 2 \cdot 2!$$ ways for the three people to return from $B$ to $A$ for each of the $5 \cdot 4 \cdot 3$ ways they could have traveled from $A$ to $B$.
Exactly two people travel from $B$ to $A$ by a route another person took to get from $A$ to $B$: There are $\binom{3}{2}$ ways to select the two people who travel from $B$ to $A$ by a route another person took from $A$ to $B$. They can either both select to return by the route the other took to get from $A$ to $B$ or exactly one of them can do so, with the other selecting the route the third person took to get there, giving $1 + 2 = 3$ choices. The third person must return using one of the two routes nobody used to travel from $A$ to $B$. Hence, there are $$\binom{3}{2} \cdot (1 + 2) \cdot 2$$ ways for the three people to travel from $B$ to $A$ for each of the $5 \cdot 4 \cdot 3$ ways they could have traveled from $A$ to $B$.
All three people travel from $B$ to $A$ by a route another person took from $A$ to $B$: There are $2$ ways this can occur for each of the $5 \cdot 4 \cdot 3$ ways they could have traveled from $A$ to $B$, namely the oldest person returns by the route the next oldest person took from $A$ to $B$, the second oldest person returns by the route the youngest person took from $A$ to $B$, and the youngest person returns by the route the oldest person took from $A$ to $B$, or the oldest person returns by the route the youngest person took from $A$ to $B$, the next oldest person returns by the route the oldest person took from $A$ to $B$, and the youngest person returns by the route the second oldest person took from $A$ to $B$.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways three people can travel from $A$ to $B$ and return from $B$ to $A$ on the five roads connecting the cities if they each take different routes from $B$ to $A$ than they took from $A$ to $B$ and no two of them take the same route in the same direction is $$5 \cdot 4 \cdot 3\left[3 \cdot 2 \cdot 2! + \binom{3}{2} \cdot (1 + 2) \cdot 2 + 2\right] = 1920$$