Okay, so the problem is about arranging medals in a row. There are two IMO gold medals and two IPhO gold medals, and all the medals are indistinguishable among themselves. The question is asking how many distinct arrangements are possible. Hmm, let's break this down step by step.

First, I need to understand what's being asked. We have four medals total: 2 IMO and 2 IPhO. Since they are indistinguishable, swapping two IMO medals or two IPhO medals doesn't create a new arrangement. So, the key here is to figure out how many unique ways we can arrange these medals in a row.

This seems like a permutations problem with repeated elements. When we have objects that are identical, the number of distinct permutations is given by the formula:

Number of permutations = n! / (n1! * n2! * ... * nk!)

Where n is the total number of objects, and n1, n2, ..., nk are the numbers of each type of identical object.

In this case, n is 4 (since there are 4 medals total), and there are two types: IMO (2 medals) and IPhO (2 medals). So, applying the formula, we get:

Number of arrangements = 4! / (2! * 2!)

Calculating that, 4! is 24, and 2! is 2, so 24 / (2*2) = 24 / 4 = 6. Therefore, there should be 6 distinct arrangements.

But let me make sure I didn't skip any steps or make a mistake here. Let me think of the possible arrangements manually to verify.

Let's denote IMO medals as 'I' and IPhO medals as 'P'. So, we need to arrange two I's and two P's in a row. The possible distinct arrangements are:

1. I, I, P, P
2. I, P, I, P
3. I, P, P, I
4. P, I, I, P
5. P, I, P, I
6. P, P, I, I

Yes, that's six different arrangements. Each arrangement is unique because the positions of I's and P's are different. For example, in the second arrangement, the first and third positions are I's, while the second and fourth are P's. This is different from the first arrangement where the first two are I's and the last two are P's. Similarly, the third arrangement has the first I, then a P, another P, and then an I. Each of these is distinct.

Another way to think about this is using combinations. Since all the medals are indistinct except for their type, we can think of choosing positions for the IMO medals and the remaining positions will be for the IPhO medals. The number of ways to choose 2 positions out of 4 for the IMO medals is C(4,2). Then, the remaining 2 positions automatically go to the IPhO medals. C(4,2) is 6, which matches the previous result.

Alternatively, using permutations: if all medals were distinct, there would be 4! = 24 arrangements. But since the two IMO medals are identical and the two IPhO medals are identical, we have to divide by the number of ways the identical medals can be arranged among themselves. For the IMO medals, there are 2! ways to arrange them, and similarly for the IPhO medals, another 2! ways. So, dividing by 2! * 2! gives 24 / 4 = 6, which again is the same answer.

Therefore, all methods converge to 6 distinct arrangements. I don't see any mistakes in this reasoning. The key was recognizing that since the medals are indistinct within their types, we need to use the formula for permutations of multiset. It's important to remember that swapping identical items doesn't create a new arrangement, which is why we divide by the factorial of the counts of each type.

So, after verifying through different methods and manually listing the possibilities, I'm confident that the answer is 6.
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