The Hawaiian language has only 12 letters: what is the probability a randomly selected 3 letter "word" The Hawaiian language has only 12 letters: the vowels a, e, i, o and u and the consonants h, k, l, m,n,p and w   what is the probability a randomly selected 3 letter "word" begins with a consonant and ends with 2 different vowels?
I did 
$$\left(\frac 7{12}\right)\left(\frac 5{11}\right)\left(\frac 4{11}\right)= \frac {140}{1320}$$ 
did I make this too simple? I keep doubting everything I am doing!
 A: Your answer would be correct if you were making selections of letters "without replacement" -- e.g., if you wrote each letter on a card, shuffled the cards, and then dealt out three of them in order.  But this isn't how random words are usually thought of, as evidenced in the constraint that the word end in two different vowels.  For the problem with replacement, the denominators on the left should all be $12$, for an answer of $(7/12)(5/12)(4/12)=140/1728=35/432$.
A: Let's first look at the total possible combinations. This is 12^3, as every space in the word can be one of 12 letters. Now we calculate the quantity of words that start with a consonant and end with two [edit: different] vowels. This is 7 * 5 * 4. So the answer would be
$$
(7 * 5 * 4) / (12^3) = 140/1728 = 0.081.
$$
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Another interesting case aside from your problem arises if we have a pool of letters where we cannot use members of the pool twice, and there are repeats of certain letters. For example, for "abb", our combinations would be:
abb, bab, bba. We always divide our final answer by the number of repeats of each letter. Thus our calculation would be 3 * 2 * 1 / 2.
So if we had to rearrange "abbccc", we make our normal calculation 6 * 5 * 4... = 720, and then divide this by repeats of a (1), repeats of b (2), and repeats of c (3), to find 720/ (1 * 2 * 3) = 120.
A: Each letter in the 3-letter word may be any of the twelve alphabets. So, the total number of cases are $n=12^3=1728$. Thus, the words aaa and hhh are also in the sample space $\Omega$. 
There are 7 consonants, so there are 7 ways to choose the first letter. There are 5 vowels, so there are 5 ways to choose the second and 4 ways to choose the third letter (since the vowels shouldn't be repeated). The number of cases favorable to the event are, therefore, $m=7\times5\times4=140$.
$P(A)=\frac{m}{n}=\frac{140}{1728}$
