Nina1987 wrote:
How is this a combination problem? See getting 10, 3, 4, 5 or 10, 4, 3, 5 or 3, 4, 5, 10 all three will give you a range of 7. So should it be a permutation problem?
It's neither a combinations nor a permutations problem; it's a probability problem.
In general, in a probability question where you are selecting things without replacement, you can pretend order matters, or pretend it doesn't, and you'll get the same answer either way. This is easier to see using a simpler question: Say you have a bag containing 7 blue marbles and 5 red marbles, and you want to know the probability you will get 2 blue marbles if you randomly pick two different marbles from the bag. You can imagine putting both hands in the bag, and grabbing two different marbles. Whether you lift your hands out one at a time (so there is a first selection and a second, and order matters) or take both hands out at once (so order does not matter), the probability you'll get two blue marbles must be the same - you're taking out the same two marbles either way. So you can answer this probability question in three ways:
- you can just use the probability rules, which is what I would do, though when we do this we're implicitly assuming we're picking in order: the probability the first marble is blue is 7/12, and then the probability the second is also blue is 6/11, so the probability both are blue is (7/12)(6/11) = 7/22
- we can use the definition of probability (which is what the solutions in this thread are using) when events are equally likely: the probability X happens = (number of ways X can happen) / (total number of things that can happen). When we use this definition for a problem like this, we can pretend order matters, or we can pretend order doesn't matter. As long as we're consistent, i.e. as long as we count the numerator and denominator under the same assumption, we'll get the right answer.
- So if we pretend order matters, we'd say "there's 7 ways for the first marble to be blue, and 6 ways for the second to be blue, and thus 7*6 ways for the first two selections to be blue, and there are 12*11 ways in total to pick two marbles in order, so the answer is 7*6/12*11 = 7/22
- Or we can pretend order does not matter. Then there are 7C2 = 7*6/2! ways to pick a set of two blue marbles, and 12C2 = 12*11/2! ways to pick a set of any two marbles, and the probability both are blue is 7C2/12C2 = (7*6/2) / (12*11/2) = 7/22
The same principle applies to the question in this thread, though I'd find this particular problem easier to solve by assuming order does not matter. Then we'd have 10C4 ways to pick four numbers. To get a range of 7, our smallest and largest numbers could be '1' and '8', and then we'd have 6C2 ways to pick our other two numbers (from the set of numbers between 2 and 7 inclusive). But our smallest and largest numbers can be 2 and 9, or can be 3 and 10, so we have 3*6C2 ways to get a range of seven, and the answer is 3*6C2/10C4 = (3)(3)(5) / (10)(3)(7) = 3/14
But you could pretend order matters instead. Then there's 10*9*8*7 ways to pick four numbers. There are going to be many ways to get a range of exactly 7. We might pick:
1, 8, X, Y
where X and Y are between 2 and 7 inclusive. Since we have 6 choices for X and 5 choices for Y, there's 30 ways to pick numbers in this precise order. But we could also pick our numbers in these orders:
1X8Y
1XY8
X18Y
X1Y8
XY18
so there are six different orders we could have of 1, 8, X and Y, for 6*30 = 180 orders, and we need to double that, because the '8' could be selected before the '1' (e.g. we could have 81XY instead of 18XY). So there are 360 ways to pick, in order, four numbers with a range of seven when the smallest selection equals 1. But the smallest selection can be 2, and can be 3, so there are three equally likely ways in total to get a range of seven, and there will be 3*360 ways to pick four numbers in order that give a range of seven. So the answer is
(3)(360)/(10)(9)(8)(7) = (3)(36)/(9)(8)(7) = 3/14
That's a correct way to solve the problem, but it's long, which is why the solutions above pretended order did not matter instead. On easier problems of this type, it's almost always easiest to assume order matters, but on harder problems, like this one, it can be easier to assume it doesn't.
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