Question 1120229
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4 + 3 + 2 = 9.


The answer to this question is:  in  {{{9!/(4!*3!*2!)}}} = {{{(9*8*7*6*5*4*3*2*1)/((4*3*2*1)*(3*2*1)*(2*1))}}} = 1260 ways.


4! in the denominator accounts for 4 red  indistinguishable disks;

3! in the denominator accounts for 3 yellow  indistinguishable disks;

2! in the denominator accounts for 2 green  indistinguishable disks.
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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.