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 $\displaystyle 9!$ 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 $\displaystyle 2!$ (permutations of S), $\displaystyle 3!$ (permutations of T) and again $\displaystyle 2!$ (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

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**MathCrusader** **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 $\displaystyle 9!$ 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 $\displaystyle 2!$ (permutations of S), $\displaystyle 3!$ (permutations of T) and again $\displaystyle 2!$ (permutations of I).

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

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

Consider $\displaystyle ABBBCD$ for our 'word'. Put subscripts in.

$\displaystyle AB_1B_2B_3CD$ now we have six different letters that can be rearranged in $\displaystyle 6!$ ways.

Here are six, $\displaystyle 3!$, of them:

$\displaystyle B_1ACB_3DB_2$

$\displaystyle B_1ACB_2DB_3$

$\displaystyle B_2ACB_1DB_3$

$\displaystyle B_2ACB_3DB_1$

$\displaystyle B_3ACB_2DB_1$

$\displaystyle B_3ACB_1DB_2$

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 $\displaystyle 3!$ to account for duplication.

Does that help.

Re: Permutation of the word STATISTIC

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

Consider $\displaystyle ABBBCD$ for our 'word'. Put subscripts in.

$\displaystyle AB_1B_2B_3CD$ now we have six different letters that can be rearranged in $\displaystyle 6!$ ways.

Here are six, $\displaystyle 3!$, of them:

$\displaystyle B_1ACB_3DB_2$

$\displaystyle B_1ACB_2DB_3$

$\displaystyle B_2ACB_1DB_3$

$\displaystyle B_2ACB_3DB_1$

$\displaystyle B_3ACB_2DB_1$

$\displaystyle B_3ACB_1DB_2$

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 $\displaystyle 3!$ 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

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**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 $\displaystyle MISSISSIPPI$ is

$\displaystyle \frac{11!}{4!\cdot 4!\cdot 2!}$. We **divide** to eliminate duplication.

**There is no subtraction to it.**

Re: Permutation of the word STATISTIC

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

The number of ways to rearrange the word $\displaystyle MISSISSIPPI$ is

$\displaystyle \frac{11!}{4!\cdot 4!\cdot 2!}$. 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

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**mathcrusader** 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.

$\displaystyle MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4$

That is eleven different letters. They can be rearranged in $\displaystyle 11!$

In **any one** of those arrangements

The **I's** represent $\displaystyle 4!$ ways.

The **S's** represent $\displaystyle 4!$ ways.

The **P's** represent $\displaystyle 2!$ ways.

So we have over-countered by $\displaystyle 4!\cdot 4!\cdot 2!$ times.

So divide.