Question 1158097
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<pre>

There are 10 letter in the word.


They all are UNIQUE, except one letter " S ", which has multiplicity of 2.


Therefore, the number of all distinguishable arrangements of the letters is  {{{10!/2!}}} = {{{3*4*5*6*7*8*9*10}}} = 1814400.    <U>ANSWER</U>


2! in the formula serves to account for repeation letter " E ".
</pre>

Solved.


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To see other similar solved problems, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Arranging-elements-of-sets-containing-undistinguishable-elements.lesson>Arranging elements of sets containing indistinguishable elements</A> 

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic &nbsp;"<U>Combinatorics: Combinations and permutations</U>". 



Save the link to this textbook together with its description


Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson


into your archive and use when it is needed.