Question number 30, the roundabout shown in the picture is entered by five cars at the same time, each one from a different direction. Each of the cars drives less than one round and no two cars leave the roundabout in the same direction. How many different combinations are there for the cars leaving the roundabout? What we need to do is keep track of the cars entering and leaving. So let's number the cars and associate a 5-tuple to the initial position. So initially, we would have 1, 2, 3, 4, 5, like so. And then after the cars drive around, we could have as an example, car 1 driving to where car 2 entered, car 2 driving to where car 4 entered. And so our position bookkeeping would look like there is a 2 in the first position and a 4 in the second position and then the other positions are to be filled in. So that's how we will keep track. And what we would do is make an observation that with five numbers, there is 120 ways to arrange that. But we will avoid answer E because we have some constraints. We have to avoid repetitions and fixed positions. If we have a repetition, that means two cars left along the same path, which is not allowed. And if we have a fixed position, that means one car made a full revolution or more and then ended up exactly where it came in. So that's also not allowed. So then what we consider are some scenarios. For example, what would happen if we had a transposition? So if cars 4 and 5 just switched positions like that, that would mean that car 4 took one turn in the direction of travel, ended up where car 5 was and car 5 would make a whole full round and turn into where car 4 came from. So that would be a transposition. So with a transposition like that, what we would have is three remaining cars and an easy way with that arrangement. So if we demand that we have a 4 and a 5 over here transposed, then let's denote it like that. If they switch positions, then what's the correct arrangement of the remaining cars? Well, 1 cannot be a 1 anymore. It can be a 2 or a 3. 2 cannot be a 2 anymore. It can be a 1 or a 3. And then 3 cannot be a 3 anymore. It could be a 1 or a 2, like that. And then we would have here a 5 and a 4, and a 5 and a 4 after transposing. So with a transposition present, there are two ways to correctly arrange the remaining three cars. So let's make a note of that. With a transposition, there are two correct combinations of the other three cars. And since we have five numbers, we can make 10 pairs or 10 different transpositions. So a total of 10 times 2 or 20 combinations. And so what happens if we don't have a transposition? Or if we have more than one transposition? Well, it's pretty easy to see that if we have two transpositions, then we run into trouble. The maximum number of transpositions allowable in our situation is just 1. And so what if we don't have a transposition? Then what happens? So I would refer you to the suggested solutions where this is considered in depth. I don't really have space here to perform that calculation. But it turns out that without a transposition, there are, in fact, 24 correct arrangements. So a total of 20 with a transposition, 24 without a transposition. So 44 ways to have the cars enter and leave according to the prescribed rules. So the answer is B.

roundabout

exit arrangements

transposition

car positions

combinations