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5 letters: a, b, c, d, e I randomly pick 3 letters, whats the chance of having a and b in those 3 letters

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You have a total of $5$ letters and you are choosing $3$ from them. Hence, the total number of ways of choosing three letters is $\dbinom{5}3$.

Number of ways in which you have $a$ and $b$ as two of your three letters is $\dbinom{3}1$, since the third letter you can choose from any of the remaining three letters $c$, $d$ and $e$.

Hence, the probability is _____.

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I presume you can also derive '3 choose 1' by saying that, for example, there is one choice for the first selection, one for the second and three for the third and so by the multi. Principle there is 3 ways. Why don't we have to divide by 3! here though if you do it this way? –  CAF Nov 24 '12 at 23:06
    
@CAF Yes. In fact that is how I have derived. However, I don't see why you want to divide by $3!$. –  user17762 Nov 24 '12 at 23:38
    
Does the method I describe not mean that I have ordered the choosing of the letters? (I.e one choice for the first letter implies this one must be 'a', for example) I think I meant multiplying by 3! neglects the order? –  CAF Nov 25 '12 at 9:19
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