What pair of numbers shows a common factor and a common multiple of 15 and 18?

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

Find the experimental probability of rolling a 4

11Alexandr11 [23.1K]

<span />the probability of rolling a 4 = 5 times.

8 0

3 years ago

A skydiver is falling at about 176 feet per second. How many feet per minute?

bixtya [17]

176 feet per sec × 6o secs in a minute = 10560

3 0

4 years ago

A population of rabbits on a farm grows by 12% each year. Assume that this increase occurs as new baby rabbits at the beginning

jolli1 [7]

Answer:

a) r₁₂ = 104.36

In general, rₙ = arⁿ⁻¹

b)

- rabbit food consumed during the 10th year is approximately 832 pounds

- rabbit food consumed in total for the 1st through 10th years is approximately 5265 pounds

Step-by-step explanation:

Given that:

r1 = 30 and a farm grows by 12%

a = 30 and the common ratio r = 1.12

now

n r

1 30.00

2 33.60

3 37.63

4 42.15

5 47.21

6 52.87

7 59.21

8 66.32

9 74.28

10 83.19

11 93.18

12 104.36

Therefore r₁₂ = 104.36

In general, rₙ = arⁿ⁻¹

b)

if each rabbit consume 10 lbs of rabbit food each year

n r food consumed(lbs)

1 30.00 300

2 33.60 336

3 37.63 376

4 42.15 422

5 47.21 472

6 52.87 529

7 59.21 592

8 66.32 663

9 74.28 743

10 83.19 832

total 5265

Therefore, the rabbit food consumed during the 10th year is approximately 832 pounds

And the rabbit food consumed in total for the 1st through 10th years is approximately 5265 pounds

6 0

3 years ago

I need help with this please help me i got it wrong

BartSMP [9]

It's a glitch I believe (screenshot and contact your teacher)

7 0

3 years ago

What pair of numbers shows a common factor and a common multiple of 15 and 18?​

What pair of numbers shows a common factor and a common multiple of 15 and 18?