Permutation of the word STATISTIC

**Q: In how many ways can the letters in the word STATISTIC be arranged?**

Now, I am not interested in the actual answer but rather the method whereby the problem is solved. We can begin by computing the number of arrangements while neglecting the repeated letters, which yields a total number of ways of arrangements.

Now here's is where I am confused: Why do we then followingly divide by the number of arrangements that can be made of the repeated letters? The right answer is obtainable by dividing by the (permutations of S), (permutations of T) and again (permutations of I).

Are we not supposed to take the overcounted arrangements and successively subtract the arrangements of the repeated letters, instead of dividing?

Re: Permutation of the word STATISTIC

Re: Permutation of the word STATISTIC

Quote:

Originally Posted by

**Plato** The best way to see is to use a simple example.

Consider

for our 'word'. Put subscripts in.

now we have six different letters that can be rearranged in

ways.

Here are six,

, of them:

Notice that the B's are in the same relative position, only the subscripts changed.

If we remove the subscripts then all six of those would be identical.

Thus the total is divided by

to account for duplication.

Does that help.

I understand your example, however I still can't we see division before subtraction is justified. We're taking a huge number 6!, i.e. the total number of arrangements, and dividing it by the arrangements of the B's i.e. 3!. We have 3! superfluous arrangements, so if one wishes to get rid of these extraneous arrangements from total number of arrangements, one would naturally want to subtract, but evidently no. Why?

Re: Permutation of the word STATISTIC

Quote:

Originally Posted by

**MathCrusader** I understand your example, however I still can't we see division **before subtraction** is justified. We're taking a huge number 6!, i.e. the total number of arrangements, and dividing it by the arrangements of the B's i.e. 3!. We have 3! superfluous arrangements, so if one wishes to get rid of these extraneous arrangements from total number of arrangements, **one would naturally want to subtract**, but evidently no. Why?

Don't you understand that there is **no subtraction** whatsoever in these problems.

The number of ways to rearrange the word is

. We **divide** to eliminate duplication.

**There is no subtraction to it.**

Re: Permutation of the word STATISTIC

Quote:

Originally Posted by

**Plato** Don't you understand that there is

**no subtraction** whatsoever in these problems.

The number of ways to rearrange the word

is

. We

**divide** to eliminate duplication.

**There is no subtraction to it.**

I do understand that, but I am still asking why since I want to deepen my understanding of this concept.

There must be rationale behind *dividing* in order to reduce the extraneous arrangements, as opposed to *subtracting* the extraneous arrangements.

Re: Permutation of the word STATISTIC