SOLUTION: If a permutation is chosen random from the letters “aaabbbccc” what is the probability that it begins with at least 2 a’s

Question 1184895: If a permutation is chosen random from the letters “aaabbbccc” what is the probability that it begins with at least 2 a’s
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

(1) Total number of permutations:

(2) Number of permutations starting with all three a's:

(3) Number of permutations starting with exactly two a's:

ANSWER: 2/21

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Tutor @ikleyn is right -- I double counted some of the permutations.

The 20 permutations in (2) above are counted again in (3); the number of permutations starting with at least 2 a's is 140, as shown in (3).

Then the probability of a permutation starting with at least 2 a's is 140/1680 = 1/12, which agrees with her answer.

Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
.
If a permutation is chosen random from the letters “aaabbbccc” what is the probability
that it begins with at least 2 a’s
~~~~~~~~~~~~~~~~~~

(1) We will consider distinguishable arrangements of 9 given letters. The total number of all such arrangements is = = 1680. (2) The number of all distinguishable arrangements starting with 3 a's is = = 20. (3) The number of all distinguishable arrangements starting with exactly 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 = 140) the number of all those distinguishable arrangements of these 7 letters, where "a" is in the first position. The latter number is = = 20. THUS, the number of all distinguishable arrangements starting with exactly 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 = = = = 0.08333... = 8.333...% . ANSWER

Solved.

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