-Find the number of distinguishable permutations of the given letters ``AAABBBCCC''.

-If a permutation is chosen at random, what is the probability that it begins with at least 2 A's?

Ah, we're talking about permutations now? This reminds me of a clown juggling balls! Let's tackle this step by step.

First, we need to find the number of distinguishable permutations of the letters "AAABBBCCC".

To do this, we calculate the total number of permutations using the formula n! / (n₁! * n₂! * ... * nₖ!), where n is the total number of letters and n₁, n₂, ..., nₖ represent the number of indistinguishable letters in each group.

In this case, we have 3 indistinguishable A's, 3 indistinguishable B's, and 3 indistinguishable C's. Plugging these values into the formula, we get:

9! / (3! * 3! * 3!)

Simplifying this expression, we find that there are 1680 different distinguishable permutations.

Now, let's move on to the second question: the probability that a randomly chosen permutation begins with at least 2 A's.

Since we have 3 A's and 6 other letters (B's and C's), there are a total of 9! permutations in total.

To calculate the number of permutations that begin with at least 2 A's, we fix those 2 A's at the beginning and count the number of arrangements for the remaining 7 letters.

This can be calculated as 7! / (3! * 2!), since there are 3 B's and 2 C's remaining.

Therefore, the probability is:

(7! / (3! * 2!)) / (9!) = 60 / 1680 = 1/28

So, the probability is 1/28, or as clowns like to say, "a slim chance, but not as slim as fitting into a mini clown car!"

To find the number of distinguishable permutations of the given letters "AAABBBCCC," we can use the concept of permutations with repetition.

Step 1: Calculate the total number of permutations disregarding any repetition. In this case, we have a total of 9 letters: 3 A's, 3 B's, and 3 C's. Therefore, the total number of permutations without any repetition is given by 9!.

Step 2: Adjust for the repetitions. Since we have 3 A's, 3 B's, and 3 C's, we need to divide the total number of permutations without repetition by the factorials of the repetition counts. So, we divide by 3! three times: (3!)^3.

Therefore, the number of distinguishable permutations is 9! / (3!)^3.

Now, let's move on to the second part of the question.

To find the probability that a randomly chosen permutation begins with at least 2 A's, we need to determine the number of favorable outcomes and the total number of possible outcomes.

Number of favorable outcomes: The starting two positions need to be occupied by A's. Since there are 3 A's, the third position can be any of the remaining 6 positions (B or C). Therefore, the number of favorable outcomes is 6!.

Total number of possible outcomes: We already calculated this in the first part of the question. The total number of distinguishable permutations is 9! / (3!)^3.

Therefore, the probability is given by the number of favorable outcomes divided by the total number of possible outcomes:

Probability = 6! / (9! / (3!)^3)

9!/(1!*3!*3!)

you take the total amount of letters you have factorial, divided by each of you different types of letters factorial. the reason that there is a 1! in the beginning is because you have the first two A's in the front, so they won't be going anywhere, leaving you with 1 A, therefore, 1!.

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