Question 1184895
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If a permutation is chosen random from the letters “aaabbbccc” what is the probability 
that it begins with at least 2 a’s
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(1)  We will consider distinguishable arrangements of 9 given letters.

     The total number of all such arrangements is  {{{9!/(3!*3!*3!)}}} = {{{392880/(6*6*6)}}} = 1680.



(2)  The number of all distinguishable arrangements starting with 3 a's is 

         {{{6!/(3!*3!)}}} = {{{720/(6*6)}}} = 20.



(3)  The number of all distinguishable arrangements starting with  <U>exactly</U>  2 a's is equal (OBVIOUSLY) 

     to the number of all distinguishable arrangements of the 7 (seven letter) "abbbccc", where "a" is not 
     
     in the first (leftmost) position.



     The number of such arrangements is equal to the number of all distinguishable arrangements of these 7 letters

     (which is  {{{7!/(3!*3!)}}} = 140)  {{{highlight(MINUS)}}}  the number of all those distinguishable arrangements of these 7 letters,

     where "a" is in the first position. The latter number is  {{{6!/(3!*3!)}}} = {{{720/(6*6)}}} = 20.



     THUS,  the number of all distinguishable arrangements starting with  <U>exactly</U>  2  a's is equal to 140 - 20 = 120.



(4)  Finally, the number of all distinguishable arrangements starting with at least 2 a's is  20 + 120 = 140.



(5)  THEREFORE, the probability under the problem's question is  P = {{{140/1680}}} = {{{14/168}}} = {{{1/12}}} = 0.08333... = 8.333...% .    <U>ANSWER</U>
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Solved.