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I have 40 elements I need to order (for simplicity just call them "1" - "40"). There are a few rules regarding how I must order some elements:

Elements "2" and "5" must go before element "10", Element "10" must go before element "7", Element "7" must go before elements "3" and "28"

How many distinct ways can I make these arrangements?

I think I have to use a partial permutation. I drew it out like this:

"2" (could have elements between) "5"...."10"...."7"...."3" (could have elements between) "28"

I'm lost after this step, however. I know how to use partial permutations with respect to preserving a group of elements that must go together but I am not sure what to make of this problem.

  • Consider inserting the $40$ numbers into an initially empty list. First there is $1$ way to insert "2". Then there are $2$ ways to insert "5", either before or after the "2". Then there is $1$ way to insert the "10", at the end. Keep inserting numbers in a manageable way. – peterwhy Feb 03 '21 at 01:09

1 Answers1

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Choose $6$ positions out of $40$, in $40\choose 6$ ways. Arrange the numbers $2,5,10,7,3,28$ into those $6$ positions so that the conditions are satisfied. This is possible in $4$ ways (you can use the order above, or you can swap $2$ with $5$ and/or $3$ with $28$). Arrange the other $34$ elements in any of the $34!$ possible ways. The total number is:

$${40\choose 6}\cdot 34!\cdot 4=\frac{4\cdot 40!}{6!}$$