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3 years ago

Determine the graph that has the following properties

Mathematics

1 answer:

uranmaximum [27]3 years ago

5 0

A. but I would double check on that one first.

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How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

3 years ago

I need some help please out

IceJOKER [234]

Question:

Solution:

Let the following equation:

$\sqrt[]{12-x}=\text{ x}$

this is equivalent to:

$(\sqrt[]{12-x})^2=x^2$

this is equivalent to:

$12-x=x^2$

this is equivalent to:

$x^2+x-12=\text{ 0}$

thus, we can conclude that

x= 3.

7 0

2 years ago

PLS HELP ASAP PLS GIVE A GOOD EXPLANATION

taurus [48]

Negative integer.

-761 + 317= -444

Plug the answer in:

-761 - (-444)
= -317

Reminder: when multiplying two negatives, you get a positive.

4 0

3 years ago

Read 2 more answers

Use Green's Theorem to calculate the circulation of F =2xyi around the rectangle 0≤x≤8, 0≤y≤3, oriented counterclockwise.

Tamiku [17]

Green's theorem says the circulation of $\vec F$ along the rectangle's border $C$ is equal to the integral of the curl of $\vec F$ over the rectangle's interior $D$ .

Given $\vec F(x,y)=2xy\,\vec\imath$ , its curl is the determinant

$\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x$