What is the formula in finding the number of distinguishable permutations of the letters of the word statistics?

There's a technique in probabilistic reasoning that is very useful in questions like this where you have "...keep $x$ fixed..." or "...at least 1 in each container...".

The trick is to first "set up" the scenario by distributing the fixed restrictions and then counting the possibilities from what you have remaining to work with. For example applied to your question you need it to start and end with an S. So from our metaphorical bag of letters, let us take out two S and place them one on each end. Now all that we need to count are the permutations of the remaining 8 letters in our bag (I'll leave this up to you).

This generalises nicely to the problem of putting $r$ objects from your collection of $n$ into $r$ particular locations leaving you with the smaller problem of just permuting $n-r$ objects.

As an other unrelated example to get my message across about the technique (because ultimately, it's the technique that you need to take away from this) consider:

Ways to distribute $n$ objects into $k$ boxes such that no box is empty. Objects indistinguishable/distinguishable and each box can hold any number of objects.

Using the technique, put exactly one object in each box and then count the number of ways to distribute $n-k$ objects into $k$ boxes. For the case where the objects are distinguishable you need to also count the number of ways to choose and permute the $k$ objects you use to fill.

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Suppose we toss a gold dollar coin 8 times. What is the probability that the sequence of 8 tosses yields 3 heads (H) and 5 tails (T)?

Two such sequences, for example, might look like this:

H H H T T T T T or this H T H T H T T T

Assuming the coin is fair, and thus that the outcomes of tossing either a head or tail are equally likely, we can use the classical approach to assigning the probability. The Multiplication Principle tells us that there are:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

or 256 possible outcomes in the sample space of 8 tosses. (Can you imagine enumerating all 256 possible outcomes?) Now, when counting the number of sequences of 3 heads and 5 tosses, we need to recognize that we are dealing with arrangements or permutations of the letters, since order matters, but in this case not all of the objects are distinct. We can think of choosing (note that choice of word!) $r=3$ positions for the heads (H) out of the $n=8$ possible tosses. That would, of course, leave then $n-r=8-3=5$ positions for the tails (T). Using the formula for a combination of $n$ objects taken $r$ at a time, there are therefore:

$\dbinom{8}{3}=\dfrac{8!}{3!5!}=56$

distinguishable permutations of 3 heads (H) and 5 tails (T). The probability of tossing 3 heads (H) and 5 tails (T) is thus $\dfrac{56}{256}=0.22$.

Let's formalize our work here!

Distinguishable permutations of $n$ objects Given $n$ objects with:

$r$ of one type, and
$n-r$ of another type

there are:

$_nC_r=\binom{n}{r}=\dfrac{n!}{r!(n-r)!}$

Let's take a look at another example that involves counting distinguishable permutations of objects of two types.

How many ordered arrangements are there of the letters in MISSISSIPPI?

Well, there are 11 letters in total:

1 M, 4 I, 4 S and 2 P

We are again dealing with arranging objects that are not all distinguishable. We couldn't distinguish among the 4 I's in any one arrangement, for example. In this case, however, we don't have just two, but rather four, different types of objects. In trying to solve this problem, let's see if we can come up with some kind of a general formula for the number of distinguishable permutations of n objects when there are more than two different types of objects.

Let's formalize our work.

number of distinguishable permutations of $n$ objects

$n_1$ are of one type
$n_2$ are of a second type
... and ...
$n_k$ are of the last type

and $n=n_1+n_2+\ldots +n_k$is given by:

$\dbinom{n}{n_1n_2n_3\ldots n_k}=\dfrac{n!}{n_1!n_2!n_3! \ldots n_k!}$

Let's take a look at a few more examples involving distinguishable permutations of objects of more than two types.

How many ordered arrangements are there of the letters in the word PHILIPPINES?

The number of ordered arrangements of the letters in the word PHILIPPINES is:

$\dfrac{11!}{3!1!3!1!1!1!1!}=1,108,800$

Fifteen (15) pigs are available to use in a study to compare three (3) different diets. Each of the diets (let's say, A, B, C) is to be used on five randomly selected pigs. In how many ways can the diets be assigned to the pigs?

Well, one possible assignment of the diets to the pigs would be for the first five pigs to be placed on diet A, the second five pigs to be placed on diet B, and the last 5 pigs to be placed on diet C. That is:

A A A A A B B B B B C C C C C

Another possible assignment might look like this:

A B C A B C A B C A B C A B C

Upon studying these possible assignments, we see that we need to count the number of distinguishable permutations of 15 objects of which 5 are of type A, 5 are of type B, and 5 are of type C. Using the formula, we see that there are:

$\dfrac{15!}{5!5!5!}=756756$

ways in which 15 pigs can be assigned to the 3 diets. That's a lot of ways!

Distinguishable Ways to Arrange the Word STATISTICS
The below step by step work generated by the word permutations calculator shows how to find how many different ways can the letters of the word STATISTICS be arranged.

Objective: Find how many distinguishable ways are there to order the letters in the word STATISTICS.

Step by step workout:

step 1 Address the formula, input parameters and values to find how many ways are there to order the letters STATISTICS.
Formula:
nPr =n!/(n1! n2! . . . nr!)

Input parameters and values:

Total number of letters in STATISTICS: n = 10 Distinct subsets: Subsets : S = 3; T = 3; A = 1; I = 2; C = 1; Subsets' count:

n1(S) = 3, n2(T) = 3, n3(A) = 1, n4(I) = 2, n5(C) = 1

step 2 Apply the values extracted from the word STATISTICS in the (nPr) permutations equation

nPr = 10!/(3! 3! 1! 2! 1! )

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10/{(1 x 2 x 3) (1 x 2 x 3) (1) (1 x 2) (1)}

= 3628800/72

= 50400 nPr of word STATISTICS = 50400 Hence, The letters of the word STATISTICS can be arranged in 50400 distinct ways.

Apart from the word STATISTICS, you may try different words with various lengths with or without repetition of letters to observe how it affects the nPr word permutation calculation to find how many ways the letters in the given word can be arranged.