Question 1121120
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<pre>
Slightly more general formulation is <U>THIS</U>:


    How many distinguishable arrangements ("words") can be formed using two letters "a" and "b".


    The answer is:  {{{5!/(2!*3!)}}} = {{{(1*2*3*4*5)/(2*6)}}} = 10.


     So you need to take 5! = 120 of all possible permutations of 5 symbols and then divide it by 2! = 1*2  and  by 3! = 1*2*3  

     to account for repeating symbols.
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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.