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The number of words that can be made by permuting the letters of *MATHEMATICS* is

$1) 5040$

$2) 11!$

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$3) 8!$

$4) 4989600$

First of all I do not understand the statement of the problem, I would like if some one tell me with an example.

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To understand the question we assume a simpler word say BALL and see how many permutations of the word exist.

BALL can be permuted as ABLL, ALBL, ALLB, BALL, BLAL, BLLA, LABL, LALB, LBAL, LBLA, LLAB and LLBA. A total of $12$ possible permutations.

The permutations of a word is given by $$\frac{\text{(Total number of alphabets)}!}{\text{(Repetitions of A)}!\text{(Repetitions of B)}!\text{(Repetitions of C)}!…\text{(Repetitions of Z)}!}$$

The number of words that can be made by permuting the letters of MATHEMATICS is

$$\frac{11!}{2!2!2!}=4989600$$

The problem is asking how many distinguishable ways you can rearrange the letters of the word MATHEMATICS. Let me give you a parallel example to see what is going on, here. Consider the word $$\text{MISSOURIANS}$$ let me rewrite it as $$\text{MIS}\color{blue}{\text{S}}\text{OUR}\color{red}{\text{I}}\text{AN}\color{green}{\text{S}}.$$ This has $11$ distinct objects in it, now that I’ve colored the repeated letters. There are $11!$ different ways we could arrange those ($11$ options for the first letter, then $10$ options for the second, then $9$ options for the third, and so on). Now, what if all of the letters letters $\text S$ were black, instead of different colors? Well, when they were different colors, each given arrangement of letters could be made into a completely different one simply by rearranging the $\text S$s (there are $3!$ ways to do this), but this makes no difference if they are all the same color. Hence, there are $$\frac{11!}{3!}$$ distinguishable ways to rearrange the letters of $$\text{MISSOUR}\color{red}{\text{I}}\text{ANS}.$$ From there, we see in a similar fashion that there are $$\frac{11!}{3!2!}$$ distinguishable ways to rearrange the letters of $$\text{MISSOURIANS}$$

See also this wikipedia article. Can you adapt this to your example?

I am describing solution.MATHEMATICS contains 11 letters so they can be arranged by $11!$ ways but in this word there is some repeating letters M is 2 times,A is 2 times, T is 2 times.So in $11!$ ways there are some permutations presents which are repeat so we have to remove those permutations that’s why divide the $11!$ by $2!\cdot 2!\cdot 2!$ (for 2 M,2 A and 2 T).suppose 1 M is on 5th place and 2nd M is on 6th place than if we swap their place the word is remain same.

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