+0

# what does the ! meaan

0
1236
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what does the ! meaan

Feb 16, 2015

#6
+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
+110206
+5

In mathematics ! is called factorial

5! means    1*2*3*4*5

* is a multiplication sign

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

Feb 16, 2015
#3
+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
+111438
+8

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

Feb 16, 2015
#5
+110206
0

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

Feb 16, 2015
#6
+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....!!!

CPhill Feb 16, 2015
#7
+110206
+5

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

Seriously now - that helped a lot :))

Feb 16, 2015
#8
+25508
+11

"

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

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

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

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

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

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

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

"

Feb 16, 2015
#9
+110206
+3

Thanks Heureka  :)))

Feb 16, 2015
#10
+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
+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