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what does the ! meaan

 Feb 16, 2015

Best Answer 

 #6
avatar+111438 
+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+110206 
+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+25508 
+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:

$$\begin{array}
$
!n = n! \cdot \displaystyle\sum^{n}_{i = 0} \frac{(-1)^i}{i!}
=n!\cdot\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots +(-1)^n \ \frac{1}{n!}
\right)$
\end{array}$$

Example !6:

$$\begin{array}
$
!6 = 6!\cdot\left(1-\dfrac{1}{1!}+\dfrac{1}{2!} -\dfrac{1}{3!} +\dfrac{1}{4!} -\dfrac{1}{5!} + \dfrac{1}{6!} \right)= 265
$
\end{array}$$

Example !7:

$$\small{\text{$
\begin{array}{rcl}
!7 &=& 7! * \left(1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+\dfrac{1}{6!}-\dfrac{1}{7!}\right)\\\\
&=& 5040*\left(\not{1}-\not{1}+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}+\frac{1}{720}-\frac{1}{5040}\right) \\\\
&=& \frac{5040}{2}-\frac{5040}{6}+\frac{5040}{24}-\frac{5040}{120}+\frac{5040}{720}-\frac{5040}{5040}\\\\
&=& 2520-840+210-42+7-1\\\\
&=&1854
\end{array}$}}$$

 Feb 16, 2015
 #4
avatar+111438 
+8

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

 

 Feb 16, 2015
 #5
avatar+110206 
0

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

 Feb 16, 2015
 #6
avatar+111438 
+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+110206 
+5

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

Seriously now - that helped a lot :))

 Feb 16, 2015
 #8
avatar+25508 
+11

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

"

$$\small{\text{Suppose that a professor has had 4 of his students }} \\
\small{\text{ - student A, student B, student C, and student D - }}\\
\small{\text{take a test and wants to let his students grade each other's tests. }}\\
\small{\text{Of course, no student should grade his or her own test. }}\\
\small{\text{How many ways could the professor hand the tests back }}\\
\small{\text{to the students for grading, }}\\
\small{\text{such that no student received his or her own test back? }}\\
\small{\text{Out of 24 possible permutations (4!) for handing back the tests, }}\\
\small{\text{there are only 9 derangements: }}\\
\boxed{BADC, BCDA, BDAC,CADB, CDAB, CDBA,DABC, DCAB, DCBA.}$$

 

$$\begin{array}{cccccr}
1& 2& 3& 4& &\small{\text{ natural Position}} \\
A& B& C& D \\
A& & & & &A \small{\text{ natural Position }} 1 \\
& B& & & &B \small{\text{ natural Position }} 2 \\
& & C& & &C \small{\text{ natural Position }} 3 \\
& & & D & &D \small{\text{ natural Position }} 4 \\
\end{array}\\\\
\small{\text{Fixed-Point-Free:}}$$

$$\\ \small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
1. \small{\text{ Derangement }} & B& A& D& C \\
B \small{\text{ not at his own(natural position) Position }} 2 & B & & & & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & A & & & \\
D \small{\text{ not at his own(natural position) Position }} 4 & & &D& & \\
C \small{\text{ not at his own(natural position) Position }} 3 & & & & C &
\end{array}
$
}} \\\\\\\\
\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
2. \small{\text{ Derangement }} & B& C& D& A \\
B \small{\text{ not at his own(natural position) Position }} 2 & B & & & & \\
C \small{\text{ not at his own(natural position) Position }} 3 & & C & & & \\
D \small{\text{ not at his own(natural position) Position }} 4 & & &D& & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & & & A &
\end{array}
$
}}$$

 

$$\\\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
3. \small{\text{ Derangement }} & B& D& A& C \\
B \small{\text{ not at his own(natural position) Position }} 2 & B & & & & \\
D \small{\text{ not at his own(natural position) Position }} 4 & & D & & & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & &A& & \\
C \small{\text{ not at his own(natural position) Position }} 3 & & & & C &
\end{array}
$
}} \\\\\\\
\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
4. \small{\text{ Derangement }} & C& A& D& B \\
C \small{\text{ not at his own(natural position) Position }} 3 & C & & & & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & A & & & \\
D \small{\text{ not at his own(natural position) Position }} 4 & & &D& & \\
B \small{\text{ not at his own(natural position) Position }} 2 & & & & B &
\end{array}
$
}}$$

 

$$\\\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
5. \small{\text{ Derangement }} & C& D& A& B \\
C \small{\text{ not at his own(natural position) Position }} 3 & C & & & & \\
D \small{\text{ not at his own(natural position) Position }} 4 & & D & & & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & &A& & \\
B \small{\text{ not at his own(natural position) Position }} 2 & & & & B &
\end{array}
$
}} \\\\\\\
\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
6. \small{\text{ Derangement }} & C& D& B& A \\
C \small{\text{ not at his own(natural position) Position }} 3 & C & & & & \\
D \small{\text{ not at his own(natural position) Position }} 4 & & D & & & \\
B \small{\text{ not at his own(natural position) Position }} 2 & & &B& & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & & & A &
\end{array}
$
}}$$

 

$$\\\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
7. \small{\text{ Derangement }} & D& A& B& C \\
D \small{\text{ not at his own(natural position) Position }} 4 & D & & & & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & A & & & \\
B \small{\text{ not at his own(natural position) Position }} 2 & & &B& & \\
C \small{\text{ not at his own(natural position) Position }} 3 & & & & C &
\end{array}
$
}} \\\\\\\
\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
8. \small{\text{ Derangement }} & D& C& A& B \\
D \small{\text{ not at his own(natural position) Position }} 4 & D & & & & \\
C \small{\text{ not at his own(natural position) Position }} 3 & & C & & & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & &A& & \\
B \small{\text{ not at his own(natural position) Position }} 2 & & & & B &
\end{array}
$
}} \\\\\\\
\small{\text{
$
\begin{array}{lcccr}
& 1& 2& 3& 4 \\
9. \small{\text{ Derangement }} & D& C& B& A \\
D \small{\text{ not at his own(natural position) Position }} 4 & D & & & & \\
C \small{\text{ not at his own(natural position) Position }} 3 & & C & & & \\
B \small{\text{ not at his own(natural position) Position }} 2 & & &B& & \\
A \small{\text{ not at his own(natural position) Position }} 1 & & & & A &
\end{array}
$
}}$$

 

$$\small{\text{In every other permutation of this 4-member set, }}\\
\small{\text{at least one student gets his or her own test back.}}$$

"

 Feb 16, 2015
 #9
avatar+110206 
+3

Thanks Heureka  :)))

 Feb 16, 2015
 #10
avatar+110206 
+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+111438 
+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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