Connected Monuments

Let the 6 monuments be $A$ , $B$ , $C$ , $D$ , (E), and $F$ . Since each monument has exactly 1 road going in and 1 road going out. A number of interconnected monuments can be regarded as a circular string. For example, for 6 Interconnected monuments, $ABDCFE$ and $BDCFEA$ are the same strings, and we have to count the amount of strings there are. To do this, we find the number of permutations there are and then divide by the number of monuments which is also the number of possible starting points of the string. This is $\frac{n!}{n} = (n-1)!$

There are a number of ways the monuments can be connected:

Case 1: 1 String only -

6 monuments all interconnected: $5! = 120$ combinations

Case 2: 2 strings -

2.1: 1 monument connected to itself, 5 interconnected: $C^6_1 0!4! = 144$ Combinations

2.2: 2 monuments interconnected, 4 monuments interconnected: $C^6_2 1!3! = 90\$$ Combinations

2.3 3 monuments interconnected, 3 monuments interconnected: $\frac{C^6_3}{2!}2!2! = 40$ Combinations

(The Expression above had to be divided by $2!$ because the two groups of 3 monuments can be reordered)

Total combinations for Case 2: $144+90+40 = 274$ Combinations

Case 3: 3 Strings -

3.1: 2 monuments interconnected, 2 monuments interconected, 2 monuments interconnected: $\frac{C^6_2C^4_2}{3!}1!1!1! = 15$ Combinations

3.2: 1 monument connected to itself, 2 interconnected, 3 interconnected: $C^6_1C^5_20!1!2! = 120\$$ Combinations

3.3: 1 monument connected to itself, 1 monument connected to itself, 4 monuments interconnected: $\frac{C^6_1C^5_1}{2!}0!0!3! = 90$ Combinations

Total combinations for Case 3: $15+120+90 = 225$ Combinations

Case 4: 4 Strings

4.1: 1 monument connected to itself, 1 monument connected to itself, 1 monument connected to itself, 3 interconnected: $\frac{C^6_1C^5_1C^4_1}{3!}0!0!0!2! = 40$ Combinations

4.2: 1 monument connected to itself, 1 monument connected to itself, 2 interconnected, 2 interconnected: $\frac{C^6_2C^4_2C^2_1}{2!2!}0!0!1!1! = 45$ Combinations

Total combinations for Case 4: $40+45 = 85$ Combinations

Case 5: 5 Strings

The Only possible combination of 5 strings to to have 2 monuments interconnected, and the rest are connected to itself. The Number of Combinations for Case 5 would be: $\frac{C^6_1C^5_1C^4_1C^3_1}{4!}0!0!0!0!1! = 15$ Combinations

Case 6: 6 strings

There is only 1 possible combination for case 6 and that is that each monument is connected to itself

In total there are: $120+274+225+85+15+1 = \boxed{720}$ Combinations

There are six options for the first monument to connect to. There are then 5 for the second, etc.

So the answer is just $6! = \boxed{720}$ .

Have I missed something?