Question 1172438
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<pre>

According to the condition, there is 1 (one) penny;  3 undistinguishable nickels; 4 undistinguishable dimes 
and 3 undistinguishable quarters.



In all, there are 1 + 3 + 4 + 3 = 11 coins.


The number of <U>distinguishable</U> arrangements is


    N = {{{11!/(3!*4!*3!)}}} = {{{(11*10*9*8*7*6*5)/(6*6)}}} = 46200,


which is exactly the number anticipated in your post.
</pre>

Solved, answered and explained.


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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.