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

Answers

Answer 1

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

[tex]\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\}[/tex]

Then

[tex]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}\}\\[/tex]

Note that since

[tex]F \cap R=\varnothing[/tex], [tex]Perm(F)\cap Perm(R)\ne \varnothing[/tex]

But since

[tex]B \cap R \ne \varnothing[/tex], [tex]Perm(B)\cap Perm(R)= \varnothing[/tex]

and

[tex]B \cap F \ne \varnothing[/tex] , [tex]Perm(B)\cap Perm(F)= \varnothing[/tex]

Since

[tex]|\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!\\[/tex]

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

and

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

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

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

This link contains another solved problem on permutations:

https://brainly.com/question/7951365


Related Questions

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Step-by-step explanation:

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Answer:

49x^4-70x^3+319x^2-420x+150, first option

Step-by-step explanation:

Just do (7x^3-5x^2+42x-30) times (7x-5)

and you will get 49x^4-35x^3-35x^3+25x^2+294x^2-210x-210x+150.

All this would simplify to the first option.

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ANSwer: The first one.

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Answers

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Answers

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Answers

Answer:

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Answers

Answer:

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Answers

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B.

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Answers

Answer:

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Answer:

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Step-by-step explanation:

8
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Answers

A: trapezoid. while trapezoids can have two right angles they are classified as having only one pair of parallel sides while the shape has two pairs

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Answers

Answer:

14

Step-by-step explanation:

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Answers

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Answers

Answer:

16093

Step-by-step explanation:

I hope this helps!

Answer:

Your answer is 6,960,212.797357.

End of Answer

Step-by-step explanation:

(19.093 * 10)^3 OR

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End of Step-by-step explanation

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Answers

Answer:

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Answers

Answer:

(2f-17h)/24k

Step-by-step explanation:

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Answers

Step-by-step explanation:

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Answers

Answer:

13/15

Step-by-step explanation:

I took the quiz and solved it out. I would give further explaination but it is kinda hard to do that in text form. Have a nice day!

The value of the expression if m = 3 is 13/15

Given the expression as shown below:

2/5 m - 1/3

If the value of m = 3, hence the expression will become;

2/5(3) - 1/3

6/5 - 1/3

Find the LCM

6/5 - 1/3 = 3(6)-5/15\

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6/5 - 1/3 = 13/15

Hence the value of the expression if m = 3 is 13/15

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Answers

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Step-by-step explanation:

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Answers

Answer:

(x^2 + 4)(x - 3).

Step-by-step explanation:

I'm guessing you want to factor this.

You can do it by grouping:

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Answers

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Hi! It’s me from your other post! Here’s my work for you! It’s been awhile since I’ve done this stuff so I’m really sorry if I get you a bad grade! I just wanted to help out:))
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