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what does the ! meaan

 Feb 16, 2015

Best Answer 

 #6
avatar+130458 
+15

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....!!!

 

 Feb 16, 2015
 #1
avatar+118696 
+5

In mathematics ! is called factorial

 

5! means    1*2*3*4*5

 

* is a multiplication sign

 Feb 16, 2015
 #2
avatar
+8

! 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

 Feb 16, 2015
 #3
avatar+26396 
+15

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!ni=0(1)ii!=n!(111!+12!13!++(1)n 1n!)$

Example !6:

!6=6!(111!+12!13!+14!15!+16!)=265$

Example !7:

!7=7!(111!+12!13!+14!15!+16!17!)=5040(11+1216+1241120+172015040)=5040250406+5040245040120+504072050405040=2520840+21042+71=1854

 Feb 16, 2015
 #4
avatar+130458 
+8

I have never heard of this before......thanks, heureka....!!!

 

 Feb 16, 2015
 #5
avatar+118696 
0

I don't understand.  What is a natural position? 

 Feb 16, 2015
 #6
avatar+130458 
+15
Best Answer

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....!!!

 

CPhill Feb 16, 2015
 #7
avatar+118696 
+5

Thanks Chris,  formulas can describe anything.  Even nut cases like ourselves :))

Seriously now - that helped a lot :))

 Feb 16, 2015
 #8
avatar+26396 
+11

see: https://en.wikipedia.org/wiki/Derangement

"

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.

"

 Feb 16, 2015
 #9
avatar+118696 
+3

Thanks Heureka  :)))

 Feb 16, 2015
 #10
avatar+118696 
+8

I have just added this thread to our reference material sticky note.

 

I do not know if other people use this but there are some really good posts in there and I use them quite often. 

 Feb 16, 2015
 #11
avatar+130458 
+13

As an addendum.....apparently......our calculator never heard of this, either...LOL!!!

If you key in  !N, where N is some positive integer........nothing.....!!!!

 

 

 Feb 16, 2015

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