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There is a set of numbers {1,2,3,4,5,6,7,8,9,10,11,12,13}. I am trying to find out how many subsets I can form that have 3 numbers in them and also the amount of subsets that will have a number 1 in them. Also, if I was to randomly generate a single outcome, what is the probability it will have a 1 in it if all outcomes have equal chance of occurring?

I know that the probability can be calculated from (Number of subsets that have 1 in them)/(Total number of permutations)

I figured out the Total total number of permutations = 13!/(13-3)! = 1716

But I can't figure out how to determine how many subsets will have a 1 in them. In addition to this, is there a way to generalise the method? For example, what if I would like to find out the number of subsets that have a number 1 and a number 2 in them for the same n and k parameters?

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  • $\begingroup$ For subsets do you mean ignoring order, which is usual, or considering order, which seems to be implied by your use of permutations? $\endgroup$ – Ross Millikan Feb 22 '18 at 15:49
  • $\begingroup$ Oh in this problem I do want to consider order. So 1,2,3 is different from 1,3,2 and so on. $\endgroup$ – Lobster Feb 22 '18 at 16:04
  • $\begingroup$ That is what I considered in my answer. You are then selecting three element permutations from thirteen, not a subset. As I commented on the other answer, the fraction that include $1$ is the same either way. $\endgroup$ – Ross Millikan Feb 22 '18 at 16:11
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It looks like instead of subsets you mean three element permutations that include $1$. You can count those by choosing first the location of the $1$, which gives three choices, then the first other element, $12$, and the second other element, $11$ for a total of $3 \cdot 12 \cdot 11=396$. You are correct that the chance a random permutation includes a $1$ is $\frac {396}{1716}=\frac 3{13}$ This is not surprising as you are choosing three elements from thirteen.

To have both $1$ and $2$ you have three places to put the $1$, two places for the $2$, and $11$ choices for the third number.

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  • $\begingroup$ Thanks! This was a great help! $\endgroup$ – Lobster Feb 22 '18 at 16:16
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This is not actually a case of permutation, but rather a case of combination. Because in a subset, the arrangement of the numbers are not important. Therefore it is the use of combinations and not permutations.

Subset 3 numbers in them = 13C3 = 286 - choosing 3 numbers out of 13

Subset 3 numbers with 1 in them = 12C2 = 66 - choosing 2 numbers given one of the number is 1 (there is only 12 numbers to pick, because 1 is chosen by default)

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  • $\begingroup$ You can use either permutations or combinations as long as you are consistent. The probability of having a $1$ comes out $\frac 3{13}$ eitherj way, as it should. There are $3!$ times more total permutations and $3!$ times more that include a $1$. $\endgroup$ – Ross Millikan Feb 22 '18 at 15:56
  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Feb 22 '18 at 16:01
  • $\begingroup$ Oh right, i forgot to answer the probability. You are absolutely right about the consistency part, and both method works. My only concern is about technicality especially with subsets, because with other question (not the probability one), the answer may be different, due to order being considered. $\endgroup$ – Zirc Feb 22 '18 at 16:02

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