Question

Find the number of distinguishable permutations of the given letters ``AAABBBCCC''. There are: ________ permutations. If a permutation is chosen at random, what is the probability that it begins with at least 2 A's? The probability is _________ .

Answer 1

There are 1680 distinguishable permutations of the letters 'AAABBBCCC'. The probability that a permutation chosen at random begins with at least 2 A's is 1/12.

Step-by-step explanation:

To find the number of distinguishable permutations of the letters "AAABBBCCC", we use the formula for permutations of a multiset: n! / (n1! × n2! × ... × nk!), where n is the total number of items to permute, and n1, n2, ..., nk are the frequencies of each distinct item.

In this case, there are 9 letters in total, with 3 A's, 3 B's, and 3 C's. Applying the formula, we have 9! / (3! × 3! × 3!) = 362880 / (6 × 6 × 6) = 1680 distinguishable permutations.

To find the probability of a permutation beginning with at least 2 A's, consider the scenarios where the first two letters are AA. There are 2 A's left and we treat the pair of A's as a single 'item' in our permutation, leading to 8 items to permute. The total number of these permutations is 8! / (2! × 3! × 3!) = 10080 / (2 × 6 × 6) = 140.

Thus, the probability is 140 / 1680 = 1 / 12.