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Three A's and seven B's are arranged at random in a row.  Find the probability that no two A's are adjacent.

 May 8, 2020
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AAABBBBBBB =10! / 3!.7! =120 permutations

 

When a permutation begins with the letter "A", there will be: 3 x 7 =21 permutations where no 2 A's will be adjacent to each other.

 

When a permutation begins with the letter "B", the first permutation will begin with:BAAABBBBBB, and each A will move 5 spots away from each other for a total of:5 x 7 "B's" =35 permutations where no 2 A's will be adjacent to each other.

 

So, the total = 21 + 35 = 56 permutations out of 120 where no 2 A's will be adjacent to each other.

 May 8, 2020
edited by Guest  May 8, 2020

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