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