Hello, in the explanation for the answer to this question Greg specified that we have 2!/2! options for the first slot, which = 1. We then complete the permutation to get 5! = 120, and we remove the duplicate O’s by then dividing by 2! to get our final answer of 60. My question is - why don’t we still have to remove the duplicate Ts as well, to get a final answer of 30? Have we essentially removed that duplicate via the 2!/2! process for this first slot? I am just confused as to how that can allow us to skip removing it later on in the process. Thanks!
Treat all the T’s and O’s as distinct, then account for repeats at the very end. Maybe that’ll make it clearer (?)
How exactly would I do that using the slot method? I thought it would be 1 option for the first slot and 5! for the remaining slots, giving us 120. But then if we account for both Ts and Os as repeats at the end we end up with 30 again.
If you fix the position of one T at the front and then place all the other letters \{O_1, O_2, M, A\}, then you automatically know where to place the other T. As such, you only really need to consider how many ways you can place those 4 letters into 5 slots while accounting for the repeated O’s.
If this doesn’t really click, you could work with a smaller example for clarity. Consider all possible arrangements of the letters TRT. If you fix one of the T’s at the start, then R can only be in the second or the third position. Now notice that once you know where the R is placed (together with the fact that we fixed one T up front), you also know where the other T must go. For example, if we place R in the second position, then the other T must be in the third position. Thus, you only need to consider how many positions R can occupy; there is no need to count cases for the other T because that calculation is already baked into everything we did prior.
Interesting. Would the method Greg originally proposed reliably work for most cases, where we remove the duplicate T’s in the first step and then only need to remove any other duplicates later on in the calculation?
Just systematically count instead of memorizing specific methods. All correct methods should converge to the same answer.
We didn’t “remove” the duplicates in the first step. Given the positions of all the other letters and one fixed T up front, you automatically know where the other T goes, so we didn’t remove anything.
Maybe I am still not completely there. So with the fixed letter case, no matter where it is fixed, if the fixed letter is one of two (e.g. 2 T’s), then we know where the other n - 2 letters can go out of n - 1 slots, so it would be P(n - 1, n - 2) and then we remove any other duplicates that aren’t the fixed letter? And so if the fixed letter was one of three (e.g. 3 T’s) then it would be n - 3 letters for n - 2 slots, and therefore P(n - 2, n - 3)? I am just trying to find a way for it to make sense to me; sorry if my logic is confusing!
Okay, let’s take a slight detour. How many distinct arrangements of ABCTT are there?
Do you see why this problem is equivalent to just counting the number of ways to distribute A, B, and C into five slots? Why is that the case? I already answered this above, but you can try to work it out as well for your own clarification/understanding.