# I can't figure out this combinatorics problem… Or at least why my solution doesn't work.

If the only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7?

So I don't understand why my solution doesn't work:

I figured if there are 4 draws, then to pick, say, $1,8$,and two numbers between $1$ and $8$, the probability would be $(1/10)*(1/9)*(6/8)*(5/7)$. You have a $1/10$ chance to pick $1$. Since there's no replacement, you have $1/9$ chance to pick $8$. then $6/8$ for integers $2,3,4,5,6,$ and $7$. Then $5/7$ for another one.

Then just multiply by three. But apparently I don't get anything close to the solution. Could someone please explain why this is?

Update: I finally get it. Thank you all for the responses!

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What is for you "the range of the numers is 7"? Or any other number instead of 7...? – DonAntonio Dec 9 '12 at 4:27
I'm not sure I understand your question, but in my solution, it would be for 1 case. The second and third cases would be 2,x1,x2,9 and 3,y1,y2,10, where x1,x2 are numbers between 2 and 9, and y1,y2 are numbers between 3 and 10. – AlanH Dec 9 '12 at 5:12
I honestly don't understand what does "the range of a set of numbers is 7, or 23, or 33445"...Does it mean they all are below the number 7, or does it mean their sum is 7...or what? Note your question doesn't deal with advanced stuff yet it hasn't received yet any answer, so perhaps other people is having a tough time trying to parse what you meant. – DonAntonio Dec 9 '12 at 5:15
Oh, there's already an answer. – DonAntonio Dec 9 '12 at 5:15
@AlanH., you're exactly right. The probabilities are the same in either scenario, but notice that you've only counted one such scenario (either 1 then 8 or 8 then 1). If, for example, you wanted to get the chance of picking a 1 and an 8 given two draws, it'd be 1/(10 choose 2) or 2(1/10)(1/9). My point was that you're missing the factor of 2 in the latter expression, which is equivalent to only counting the 1-8 ordered scenario. – Josh Keneda Dec 19 '12 at 0:39

There are $\dbinom{10}{4}$ equally likely ways to pick $4$ numbers. The number of ways to pick $1$, $8$, and two from the $6$ numbers from $2$ to $7$ inclusive is $\dbinom{6}{2}$.

Then multiply by $3$.

Remark: In your solution, implicitly the numbers are being obtained in some specific order, so the probability obtained is much too low. One can correct for this by multiplying by the number of permutations that were double counted. That can be done, though the right factor may not be obvious.

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tofu, you have found the probability that you pick $1$, then $8$, then two others between $1$ and $8$. But you could also have picked $8$, then $1$, then the other two. Or one of the other two, then $1$, then $8$, then the other of the other two. So what you need to figure out is how many orderings there are of one-eight-other-other, and multiply by that (and then by $3$).

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