I think he means this, Melody...
Let's suppose that we have the ordered elements of a set, {A,B,C}
Then, the possible ways that none of them can be in their "natural" positions is just
{B,C,A} and {C,A,B} = 2 "derangements"
And this equals.....according to heureka's formula....
3! ( 1 - 1/1! + 1/2!+ 1/3!) =
3! ( 1/2! + 1/3!) =
3! [3 - 1]/ 6=
6 (2/6] =
6[1/3] =
2 "derangements"
Hey.....I've been "deranged" my whole life.....now .....I've found the "formula" as to why....!!!
In mathematics ! is called factorial
5! means 1*2*3*4*5
* is a multiplication sign
! is a factorial. To take the factorial of a number would mean to multiply it so that it multiplies itself, along with each number before it in descending order.
So 100! Would be 100 x 99 x 98 x 97 x 96 x 95 x 94 x 93 and so on until you get to 0
What does the ! mean ?
The subfactorial or left factorial, written, is the number of ways that n objects can be arranged where no object appears in its natural position (known as "derangements.")
The formula:
!n=n!⋅n∑i=0(−1)ii!=n!⋅(1−11!+12!−13!+⋯+(−1)n 1n!)$
Example !6:
!6=6!⋅(1−11!+12!−13!+14!−15!+16!)=265$
Example !7:
!7=7!∗(1−11!+12!−13!+14!−15!+16!−17!)=5040∗(⧸1−⧸1+12−16+124−1120+1720−15040)=50402−50406+504024−5040120+5040720−50405040=2520−840+210−42+7−1=1854
I think he means this, Melody...
Let's suppose that we have the ordered elements of a set, {A,B,C}
Then, the possible ways that none of them can be in their "natural" positions is just
{B,C,A} and {C,A,B} = 2 "derangements"
And this equals.....according to heureka's formula....
3! ( 1 - 1/1! + 1/2!+ 1/3!) =
3! ( 1/2! + 1/3!) =
3! [3 - 1]/ 6=
6 (2/6] =
6[1/3] =
2 "derangements"
Hey.....I've been "deranged" my whole life.....now .....I've found the "formula" as to why....!!!
Thanks Chris, formulas can describe anything. Even nut cases like ourselves :))
Seriously now - that helped a lot :))
see: https://en.wikipedia.org/wiki/Derangement
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Suppose that a professor has had 4 of his students - student A, student B, student C, and student D - take a test and wants to let his students grade each other's tests. Of course, no student should grade his or her own test. How many ways could the professor hand the tests back to the students for grading, such that no student received his or her own test back? Out of 24 possible permutations (4!) for handing back the tests, there are only 9 derangements: BADC,BCDA,BDAC,CADB,CDAB,CDBA,DABC,DCAB,DCBA.
1234 natural PositionABCDAA natural Position 1BB natural Position 2CC natural Position 3DD natural Position 4Fixed-Point-Free:
12341. Derangement BADCB not at his own(natural position) Position 2BA not at his own(natural position) Position 1AD not at his own(natural position) Position 4DC not at his own(natural position) Position 3C 12342. Derangement BCDAB not at his own(natural position) Position 2BC not at his own(natural position) Position 3CD not at his own(natural position) Position 4DA not at his own(natural position) Position 1A
12343. Derangement BDACB not at his own(natural position) Position 2BD not at his own(natural position) Position 4DA not at his own(natural position) Position 1AC not at his own(natural position) Position 3C 12344. Derangement CADBC not at his own(natural position) Position 3CA not at his own(natural position) Position 1AD not at his own(natural position) Position 4DB not at his own(natural position) Position 2B
12345. Derangement CDABC not at his own(natural position) Position 3CD not at his own(natural position) Position 4DA not at his own(natural position) Position 1AB not at his own(natural position) Position 2B 12346. Derangement CDBAC not at his own(natural position) Position 3CD not at his own(natural position) Position 4DB not at his own(natural position) Position 2BA not at his own(natural position) Position 1A
12347. Derangement DABCD not at his own(natural position) Position 4DA not at his own(natural position) Position 1AB not at his own(natural position) Position 2BC not at his own(natural position) Position 3C 12348. Derangement DCABD not at his own(natural position) Position 4DC not at his own(natural position) Position 3CA not at his own(natural position) Position 1AB not at his own(natural position) Position 2B 12349. Derangement DCBAD not at his own(natural position) Position 4DC not at his own(natural position) Position 3CB not at his own(natural position) Position 2BA not at his own(natural position) Position 1A
In every other permutation of this 4-member set, at least one student gets his or her own test back.
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