Question 1160676
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<pre>

As the condition says, the store has two digits { 3 ";  one digit  " 5 ", and 3 digits  " 6 ".


So, in all there are 6 digits at the store;

of them, digit  3  has the multiplicity of 2;

         digit  5  has the multiplicity of 1;

    and  digit  6  has the multiplicity of 3.


The number of all possible different six-digit arrangements of these digits is


    {{{6!/(2!*3!)}}} = {{{(6*5*4*3*2*1)/(2*3)}}} = 120.


In the formula, 6! is the number of all possible permutations of 6 different digits;

the factors in the denominator account for repeating arrangements, that occur due to presence of indistinguishable digits.
</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.


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.