What is distinct letters in the word mathematics

Note: This answer presumes that the question is "How many four-letter words can be formed using the letters of the word 'MATHEMATICS'." So MAME should be counted as a word.

The provided answer, ${}_{11}P_4 / (2!)^3$, is not correct.

Let's temporarily change the question to "How many distinct rearrangements are the of the letters of MATHEMATICS?", that is, how many 11-letter words can we form. If the letters were all distinct, the answer would be 11!, that is, ${}_{11}P_11$. But the letters are not all different, so we must adjust. We can do so as in the original problem: First, attach subscripts to identical letters, so that the letters are in fact all distinct, making 11! "words".

Now we need to erase subscripts. First take the Ms. There are two of them, M1 and M2, and swapping these does not change a "word" when we're ignoring subscripts. So there are 11!/2 words if we ignore the subscripts on M. Proceeding similarly, we find that there are $11!/(2*2*2)$ distinct words when all the subscripts are erased.

We need factorials when we have more than two of a letter. Suppose there were three Xs. Then there are 3! ways of rearranging the Xs, and all these ways lead to the same word when subscripts are erased. So the adjustment would be to divide by 3!. (Really this is happening with the Ms too, since 2! = 2.)

In general: Suppose we have a total of $n$ letters, including $c_1$ copies of one letter, $c_2$ copies of the second letter, . . . , and $c_i$ copies of the $i^{\rm th}$ letter. Then the total number of words we can make is

$$n! \over { c_1!c_2!\ldots c_i! }$$

Notice that this formula works even for letters that aren't replicated, since 1! = 1.

So far so good. The trouble is that the original problem does not ask for words formed from all the letters, just from four at a time. And here the trick doesn't work. We do need to adjust, but we can't simply divide by 2! when there are two copies of a letter, because some of the words don't contain all the copies of that letter! The adjusting trick, as described, works only when all copies of the letter are necessarily in the word, i.e., when each word contains all letters.

To see this even more clearly, suppose I had asked for words of two letters formed from the letters of MATHEMATICS. Would the answer be ${}_{11}P_2/(2!)^3$? Certainly not, since this expression is not even an integer.

Is there a way of phrasing the question so that the provided answer is correct? I don't see it, and I find it doubtful anyway, for the same reason as in the preceding paragraph.

Answer

Hint: First find the number of ways in which word ‘Mathematics’ can be written, and then we use permutation formula with repetition which is given as under,Number of permutation of $n$objects with$n$, identical objects of type$1,{n_2}$identical objects of type \[2{\text{ }} \ldots \ldots .,{\text{ }}{n_k}\]identical objects of type $k$ is \[\dfrac{{n!}}{{{n_1}!\,{n_2}!.......{n_k}!}}\]

Complete step by step solution:

Note: In this type of question, we use the permutation formula for a word in which the letters are repeated. Otherwise, simply solve the question by counting the number of letters of the word it has and in case of the counting of vowels, we will consider the vowels as a single unit.

Distinguishable Ways to Arrange the Word MATHEMATICS
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 MATHEMATICS be arranged.

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

Step by step workout:

Input parameters and values:

n1(M) = 2, n2(A) = 2, n3(T) = 2, n4(H) = 1, n5(E) = 1, n6(I) = 1, n7(C) = 1, n8(S) = 1

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

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11/{(1 x 2) (1 x 2) (1 x 2) (1) (1) (1) (1) (1)}

= 39916800/8

Apart from the word MATHEMATICS, 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.

The number of distinct arrangements of the letters of the word BOXING is the same as the number of permutations of 6 things taken 6 at a time. This is 6 factorial, which is 720. Since there are no duplicated letters in the word, there is no need to divide by any factor.