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Suppose you are given the digits 1, 1, 2, 2, 3, 3, 4, 4. How would you find the number of 3-digit permutations you can create from those?

For 8-digit permutations the answer should be $\frac{8!}{2!2!2!2!}$ I believe.

However, I cannot think of a way to do this when asked for permutations where not all of the set is used. I have tried $\frac{8P3}{2!2!2!2!}$ but that does not match the amount when I try manually listing all permutations which seems to be 60.

If the problem were to be generalized what would a solution look like?

Thanks in advance.

combperm
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  • I'd go by cases. Either you are permuting three distinct numbers or you are permuting $X,X,Y$ where $X\neq Y$. Both cases are easily counted. – lulu Sep 03 '19 at 13:46
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    Hint: How many are there if you have available infinitely many $1$'s, infinitely many $2$'s, etc...? In other words, how many length three strings do you have where you don't care about if you "used too many" of any particular digit? Now... how many of these were "bad" and we didn't actually want to count them for our original problem? – JMoravitz Sep 03 '19 at 13:46
  • As for the fully generalized problem where the number of types of available digits may vary, the number of available digits of each type may vary, and the length of the string we are wishing to create may vary as well... the final formula would be incredibly ugly and tedious to write out, but the easiest approaches would usually be with either inclusion-exclusion or generating functions. – JMoravitz Sep 03 '19 at 13:52
  • @JMoravitz Thank you for your comments. I'm wondering though about this website link. What is the difference between their example and mine such that their formula does not work for my case. – combperm Sep 03 '19 at 14:13
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    Reading the page again more closely... The formula they propose supposedly counts the number of $5$ letter words from the word DEFINITION... the formula is flat out incorrect... This should be obviously incorrect because taking the proposed formula to the extreme, counting how many $1$ letter words can be formed... the answer is obviously not $\frac{10}{3!\cdot 2!}$ as that is not even an integer. – JMoravitz Sep 03 '19 at 14:23
  • The page I linked this to explains the generating function approach and gives a correct answer to the problem in your link. The generating function approach is incredibly useful for the generalized problem, but for your original question here, I still say inclusion-exclusion was the easier one to use. To finish off that hint, there are $4^3=64$ possible three-digit numbers you could have made, but $4$ of them were bad since they used too many of a digit (namely 111,222, ...). Subtracting those we are left with $64-4=60$ good arrangements. – JMoravitz Sep 03 '19 at 14:31
  • @JMoravitz Thank you for your explanations, and for the clarification around the link – combperm Sep 03 '19 at 14:35
  • To emphasize, the number of five letter arrangements using letters from DEFINITION is not $\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{3!2!}=2520$ as they suggest, but is rather $5380$. As another example for why the proposed formula on their site is wrong, take your original question here but look at the number of two-digit numbers that can be made. $\frac{8\cdot 7}{2!2!2!2!}$ is not even an integer and cannot be the answer to a counting question. – JMoravitz Sep 03 '19 at 14:37

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