# Math Help - Combinations of the letters in 'statistics'?

1. ## Combinations of the letters in 'statistics'?

I've just worked it out for permutations but i'm struggling to get the right answer for combinations.

2. ## Re: Combinations of the letters in 'statistics'?

Hi Wevans2303!

What did you try?
How many permutations did you get?

3. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by ILikeSerena
Hi Wevans2303!

What did you try?
How many permutations did you get?
I got 10!/(3!3!2!) = 50400 for permutations.

For combinations I don't know the general formula so I guessed at 10!/(3!3!2!)(10!-8!) but that was incorrect.

4. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by Wevans2303
I got 10!/(3!3!2!) = 50400 for permutations.

For combinations I don't know the general formula so I guessed at 10!/(3!3!2!)(10!-8!) but that was incorrect.
I think you're mixing up permutations with combinations.
The number of permutations is 10!
The number of combinations is what you already have...

5. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by ILikeSerena
I think you're mixing up permutations with combinations.
The number of permutations is 10!
The number of combinations is what you already have...
No I checked it and I am correct.

See Stats: Counting Techniques

6. ## Re: Combinations of the letters in 'statistics'?

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?

7. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by ILikeSerena
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?

If I check it on wolfram, I get 50400 permutations and 732 combinations.

permutations &#123;s,t,a,t,i,s,t,i,c,s&#125; - Wolfram|Alpha

combinations &#123;s,t,a,t,i,s,t,i,c,s&#125; - Wolfram|Alpha

8. ## Re: Combinations of the letters in 'statistics'?

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?

9. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by ILikeSerena
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)...?

10. ## Re: Combinations of the letters in 'statistics'?

Hello, Wevans2303!

Can you give us the original wording of the problem?

Number of permutations of the letters in STATISTICS ?
The answer is: . $\frac{10!}{2!\,3!\,3!} \:=\:50,\!400$

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!

11. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by Soroban
Hello, Wevans2303!

Can you give us the original wording of the problem?

The answer is: . $\frac{10!}{2!\,3!\,3!} \:=\:50,\!400$

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!

Hi, yes I agree the wording of the question is a little ambiguous, it just says 'combinations' but I am 99% certain it means distinct subsets.

It just says 'find the number of combinations of the letters in statistics'.

12. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by Soroban
Wolfram's solution (not explained clearly) seems to provide:
. . the number of distinct subsets of the ten letters.
Hah! Not even that, since {s,s} is counted as a separate combination, which is not a "subset".

I do wonder what kind of definition Wolfram uses, since I don't know it (yet).

13. ## Re: Combinations of the letters in 'statistics'?

Originally Posted by Wevans2303
My lecturer mentioned something like:

(n1r1)(n2r2)(n3r3)...?
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.