Question 1176493
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<pre>

There are 11 symbols/letters in all.


Of them, "I" has the multiplicity 4;

         "S" has the multiplicity 4;

         "P" has multiplicity 2.


Therefore, the number of all possible distinguishable arrangements of the letters is   {{{11!/(4!*4!*2!)}}} = {{{(11*10*9*8*7*6*5*4*3*2*1)/(24*24*2)}}} = 34650.


The factorials in the denominator account for permutations of repeating letters.
</pre>

Solved, answered and explained.


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If you need more explanations or if you want to see many other similar solved problems, look into this 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.