I've just worked it out for permutations but i'm struggling to get the right answer for combinations.
Any advice appreciated.
No I checked it and I am correct.
See Stats: Counting Techniques
If you read carefully, you'll see that it is called the number of "distinguishable permutations" which is different from the number of permutations.
Either way, I interpreted your problem statement to ask for the number of distinguishable permutations.
If we look only at the combinations of all of the letters of STATISTICS, there is only 1.
Can you clarify which combinations are intended?
I'm not entirely sure what you mean, sorry about this..
If I check it on wolfram, I get 50400 permutations and 732 combinations.
permutations {s,t,a,t,i,s,t,i,c,s} - Wolfram|Alpha
combinations {s,t,a,t,i,s,t,i,c,s} - Wolfram|Alpha
All right, I give up!
I'd rather not go into hairsplitting about the meaning of words anyway.
So let's focus on the combinations.
Wolfram shows the actual combinations, which are all unordered letter combinations of zero or more letters.
This is not so easy to calculate afaik.
Wolfram appears to do it by simply enumerating all possibilities.
Are you supposed to use a calculator to do this?
Or are you supposed to find a formula for it?
I'll have to think about it a bit more to find a generic formula...
Edit: This is supposed to be "Basic Statistics and Probability" isn't it?
Well thank you for being patient. I assumed it was basic and I was missing something obvious, as the permutations formula is not that complex.
I'm expected to know the formula and manipulate the factorials.
My lecturer mentioned something like:
(n1r1)(n2r2)(n3r3)...?
Hello, Wevans2303!
Can you give us the original wording of the problem?
The answer is: .$\displaystyle \frac{10!}{2!\,3!\,3!} \:=\:50,\!400$Number of permutations of the letters in STATISTICS ?
As ILikeSerena pointed out there is one combination for the ten letters.
Wolfram's solution (not explained clearly) seems to provide:
. . the number of distinct subsets of the ten letters.
If that's what the problem wanted, it should have said so . . . clearly!
Well, let's see... if we just look at the combinations of 3 letters, we get:
1: sss
1: ttt
(4 1): ss <other letter>
(4 1): tt <other letter>
(4 1): ii <other letter>
(5 3): 3 different letters
So the number of combinations of 3 letters out of "statistics" is:
1+1+4+4+4+10 = 24
We would have to repeat something like this for each number between 0 and 10.
No. I'm not getting a simple formula yet.