#6**+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....!!!

CPhill
Feb 16, 2015

#1**+5 **

In mathematics ! is called factorial

5! means 1*2*3*4*5

* is a multiplication sign

Melody
Feb 16, 2015

#2**+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

Guest Feb 16, 2015

#3**+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}$}}$$

heureka
Feb 16, 2015

#6**+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**+5 **

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

Seriously now - that helped a lot :))

Melody
Feb 16, 2015

#8**+10 **

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.}}$$

"

heureka
Feb 16, 2015