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Sum-thing else

In addition to the recursion formulas shown, a subfactorial  can be developed using a summation. As an example, we'll develop the result for !4 , which we have addressed in an earlier post. ...which is the same answer we developed using the recursion  approach.  
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Is there a relationship between n! and !n?

 Yes.   Oh, you want more. Right, then: and (even cooler [if that's possible!]!)  

Seriously...I don't have time to click the links on this page. What's "subfactorial"?

We're glad that you asked. The n th subfactorial is the number of permutations of n objects in which no object appears in its natural place (from Wolfram). So, if we have  the digits {1, 2, 3, 4} ( n=4), we want to know how many ways can we order those numbers such that none  appear in their initial position (so, "1" will not be first; "2" will not be second, "3" will not be third, and "4" will not be fourth). The following combinations of {1, 2, 3, 4} are the only combinations in which no digit is in its natural place: {2, 1, 4, 3} {2, 3, 4, 1} {2, 4, 1, 3} {3, 1, 4, 2} {3, 4, 1, 2} {3, 4, 2, 1} {4, 1, 2, 3} {4, 3, 1, 2} {4, 3, 2, 1} There are 9 possible combinations that meet this criterion. So !4=9 . Remember that there are 4! (=24) possible combinations of {1, 2, 3. 4}, so we note that 4!-!4=24-9=15  of them have a value that is  in its natural place. The following is a list of combinations that do not  meet the criterion (that ...