Question 1148680
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<pre>

In all, there are 11 letters in the word.


Of them, letter "E" has multiplicity 3;

         letter "N" has multiplicity 3;

         letter "G" has multiplicity 2;
 
         letter "I" has multiplicity 2;

         the last letter "R" has multiplicity 1.


The number of distinguishable permutations of letters is  {{{11!/(3!*3!*2!*2!)}}} = {{{(11*10*9*8*7*6*5*4*3*2*1)/(6*6*2*2)}}} = 277200.    <U>ANSWER</U>
</pre>

Solved.


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See 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.