Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 0 down vote favorite
1

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

share|improve this question

1 Answer

up vote 1 down vote accepted

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 _____.

share|improve this answer
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.