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Two numbers are randomly selected from the set $S=\{1,2,3,4,5,6\}$ without replacement one by one. The probability that minimum of the two numbers is less than $4$ is?

As we have to find out the probability here, I noticed that the arrangements cancel out. So the answer came out to be the same even if arrangements were not considered. So, are arrangements of the numbers important here?

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    Hint: if the minimum exceeds (or equals) $4$ then the two were drawn from $\{4,5,6\}$ – 2017-02-13
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    Note: I don't understand your question about arrangements. The minimum is the same whether you consider order or not. – 2017-02-13
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    I understand. But should I consider the arrangements? Or are the cases(1,3),(3,1) taken to be the same? – 2017-02-13
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    If I understand correctly, and firstly assuming that their is replacement of numbers, wouldn't you be able to create a $6 \times 6$ array where each element is equal to the sum of the row and column reference. Out of the $36$ elements there would be $6/36=1/6$ probability. If there is no replacement then the array would be upper (lower) triangular. – 2017-02-13
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    That is entirely up to you. do it both ways, you'll get the same result. – 2017-02-13
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    Yes, but if we didn't have to find the probability if we had to find the no.of ways? – 2017-02-13
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    The number of ways to do what? If you were asked for the number of ways to draw one number after the other, then order matters as it is part of the question. – 2017-02-13
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    Yes but the minimum number is less than four in both these cases(1,2),(2,1) why do they have to be taken as different? – 2017-02-13
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    As I said, they don't have to be taken as different. That's a choice you make. – 2017-02-13
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    As a general rule: if there is doubt, include the order. Sometimes it matters (but not here). If, for example, we are considering the toss of a pair of fair dice then, as unordered pairs, you don't get equiprobable events. That is, $\{1,1\}$ is less probable than $\{1,2\}$ because there are two ways to get the latter. If you reset your problem to include replacement, order would matter for this reason. – 2017-02-13

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If $k$ numbers are chosen without replacement, every set of $k$ numbers has the same number of arrangements, $k!$. So for this sort of question you can ignore the arrangements without affecting the probability. One way you say there are $12$ pairs of numbers with minimum less than $4$, and $15$ pairs total, so the probability is $4/5$. The other way you say there are $24$ arrangements with minimum less than $4$, out of $30$ arrangements in total, so the probability is $4/5$. This works because all the possible arrangements are equally likely and all possible sets are equally likely.

If the numbers are chosen with replacement, say by rolling a die, not every collection of numbers has the same number of arrangements. In this case it is vital to count arrangements, because it is the arrangements (not the possible collections) which all have the same probability. So with replacement there are $27$ arrangements out of $36$ which work, so $p=3/4$. It would be wrong to say there are $15$ collections of numbers which work, out of $21$ possible collections, because these collections are not equally likely (it is easier to roll a 1 and a 2 in some order than two 1s).