Question 1155407
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<pre>


The word "missouri" has 8 letters, of them letter "i" has a multiplicity of 2 and the letter "s" also has a multiplicity of 2.


Therefore, the number of all distinguishable permutations under the question is


    n = {{{8!/(2!*2!)}}} = {{{(8*7*6*5*4*3*2*1)/(2*2)}}} = 10080.


Two factors  2!  in the denominator serve to account for multiplicities of the two letters, "i" and "s".
</pre>

Solved.


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To see many other similar solved problems, 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.