S1 Permutations Combinations

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Is this really S1? And have you given the question exactly as worded in the source document?

Yes to both. It's an S1 extension question my teacher said we needed to figure out how to do.

I'm aware that this requires both nCr and nPr along with duplication modifications. I just don't know how to do that yet.

Never heard of S1 extension.

I'd start by counting the number of each letter.

Then one method would be to break it down into different cases depending on the duplication factors.

E.g no. of choices such that you have 4 of one letter, 4 of another, and 1 of something else.

By my reckoning there are 18 cases to consider, and some of them quite complex to evaluate. You'll need to be methodical.

Rather tedious, so makes me wonder if there is a simpler method; perhaps involving polynomials?

How about this for the combinations:

n=24, r=9, d=3! (I) * 4! (S) * 2! (E) * 2! (T) * 3! (A) * 2! (M) * 2! (N)

And then just plug into my general formula (I have no clue whether it will work for this, though it did solve a simpler example I looked at earlier today):

N=((n!)/(d!))/(nCr) where d! is the correction for repetition.
N=3.4326*10¹³.

So would you say my general formula is correct?

Don't follow what you're doing there.

However, if all the letters were different there would be 24C9 = 1307504 combinations. Since there is duplication the actual number will be less than that, so what you have can't be right.

Edit: Looking at it in terms of polynomials, if I got it right,

OK, it's probably better we leave that method aside.

How would I do it if rather than finding combinations of 9 letters I had to find combinations of all 24? (Given that there is some duplication.) That is easier I'm sure. e.g. does n just fall to the number of distinct elements and then (n-d)Cr=(n-d)C(n-d)=1?