Let's imagine that instead of writing out a sequence of letters, we're pairing each letter with a number between 1 and 26. This is mathematically equivalent, but it'll make talking about it easier.
## of Permutations
graceful bane
Now, the first thing to note is that we want arrangements where a, b < c < d < e, f
We can simplify this by selecting arrangements where a < b < c < d < e < f
and multiplying our final solution by four.
The second thing to note is that we don't care at all about the positions of 20 of the letters, which means we can just focus on our six letters and then multiply the number of arrangements we find for that by 20!.
tough girder
Now, the first thing to note is that we want arrangements where```
a, b < c < d ...
a < b < c < d < e < f works, but so does a < b < c < d < f < e
Hmm. Wait, are there only 5 possible permutations then?
graceful bane
a < b < c < d < e < f works, but so does a < b < c < d < f < e
...I know, I just accounted for that.
I said this was a simplification and how to correct for it.
graceful bane
Yeah.
tough girder
Yeah, okay, that makes sense. I'm sorry lol
Is the answer literally just 20! * 4 then?
graceful bane
No, it's not.
tough girder
c must go in the middle, and d must go before e and f, so that restricts d to that fourth spot, and e and f to those final two spots right?
likewise for a,b?
graceful bane
...here's how I think about it.
Any time you're dealing with permutations, doing some casework can help you build an intuition about the patterns involved.
So let's consider the case where (a, b, c, d, e) = (1, 2, 3, 4, 5). How many positions can f occupy in this case?
tough girder
So let's consider the case where `(a, b, c, d, e) = (1, 2, 3, 4, 5)`. How many p...
2 places
tough girder
1, 2, 3, 4, 5, f and 1, 2, 3, 4, f, 5?
tough girder
...explain.
Are you saying that once you set e, then f can only occupy one other place?
And likewise for f...
if you set f, then e can only occupy a singular space
so setting e is effectively setting f. so it's just counting how many ways you can set a and e? (namely, two ways for both) so 4 total ways?
graceful bane
1, 2, 3, 4, 5, f and 1, 2, 3, 4, f, 5?
...what?
tough girder
(a, b, c, d, e, f) = (1, 2, 3, 4, 5, f)
graceful bane
You can either switch e with f, or keep f on the last position...
...the correct answer is 21. There are 21 possible places for f.