Question 1210204
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the number of ways of arranging one A, two Bs, three Cs, and four Ds, so that 
no two Bs are next to each other, no two Cs are next to each other, and no two Ds are next to each other.
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Under the link 
<A HREF=https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1210250.html>https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1210250.html</A>


at this forum I solved similar problem


<pre>
    Find the number of ways to arrange 4 green balls, 3 red balls, and 2 white balls 
    in a straight line such that no two balls of the same color are adjacent to each other.
</pre>


The answer to this problem was that the number of ways is 79.


In other terms, there are 79 words of the length 2+3+4 = 9 letters/positions, consisting of two letters Bs,
three letters Cs and four letters Ds so that no two Bs are next to each other, no two Cs are next to each other, 
and no two Ds are next to each other.


I will not repeat this solution here: it is quite long.
Simply take this result as is.


Now, having one additional letter A, for our problem, we can (and we should) place it in any position 
of these 79 words of the length 9, without any constraints.


Doing it, we will obtain 10 different words of the length 10 for each of the previous words of the length 9.


Hence, in all, we will have 10*79 = 790 of the words of the length 10, consisting of one A, two Bs, three Cs, 
and four Ds, so that no two Bs are next to each other, no two Cs are next to each other, and no two Ds 
are next to each other.


At this point, the problem is solved completely, and the answer is


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;there are 790 such words, or 790 such arrangements,  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;satisfying the problem's conditions.



Solved in full.



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You can ignore the post by @CPhill, since his approach is incorrect and does not lead to the solution.