Question

I AM GIVING 45 POINTS TO WHOEVER GETS THIS RIGHT... plz answer corectly and try... i was working on this for sooo long. Plz try your hardest :(
A 6-letter "word" is obtained by spinning a fair spinner (labeled a,b,c) six times. What is the probability of the word containing two of each of the three letters? What is the probability of the word containing no b's and at least two a's?

Answer 1

When you have 3 choices for each of 6 spins, the number of possible "words" is
  3^6 = 729

The number of permutations of 6 things that are 3 groups of 2 is
  6!/(2!×2!×2!) = 720/8 = 90

A) The probability of a word containing two of each of the letters is 90/729 = 10/81


The number of permutations of 6 things from two groups of different sizes is
  (2 and 4) : 6!/(2!×4!) = 15
  (3 and 3) : 6!/(3!×3!) = 20
  (4 and 2) : 15
  (5 and 1) : 6
  (6 and 0) : 1

B) The number of ways there can be at least 2 "a"s and no "b"s is
  15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.


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These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.