Permtations/Combinations Etc (1 Viewer)

[Kutay]

Hey, i have a question on a couple of questions,

"In a random arrangement of the letters of the word NUMBER what is the probabilty that the two vowels will come together?"

"How many numbers divisible by 5 can be made with the digits 2, 3, 4 or 5 if each digit is not used more than once?"

Just wondering if you can show how to answer the following questions, as i got stuck thank you

Also i just want to clarify a answer which i didn't get to one question but i fill i am correct!!!

Question is "in how many ways can the lettters of the word READER be arranged?"

For this question i did this as

6!
-----
2! x 2!

the 2! x 2! Is because of the two times that "r" is used and "e"!!! and i get an answer of 180, however in the answers it says its 360????? can anyone confirm my thoughts on the answers being wrong or if its just me,

Thank You

 

[Dreamerish*~]

"In a random arrangement of the letters of the word NUMBER what is the probabilty that the two vowels will come together?"

UE NMBR

Since UE is together, there are 5 objects therefore the number of arrangements is 5!. But since it could be UE or EU, it's 5! x 2 = 240.

The total number of arrangements is 6! = 720. So the probability of two vowels being together is 240/720 = 1/3.

"How many numbers divisible by 5 can be made with the digits 2, 3, 4 or 5 if each digit is not used more than once?"

This means 5 occupies the last digit. So basically it's how many arrangements of 2, 3, and 4 there are when the last digit has already been decided. So it's 3! = 6. (This is only if the number has to be 4-digit. For the correct answer, see my post below )

For the last one, I think you're right. It's 180. I think whoever wrote the answer missed a 2!.

 

[Kutay]

COOL thankyou so much for all the help, your a lifesaver,

Yeh i think the book is wrong!

 

[100percent]

"In a random arrangement of the letters of the word NUMBER what is the probabilty that the two vowels will come together?"

total arrangements = 6! => 720
vowels together arrangements = 5!x2! => 240
probability 360/720 => 1/3

"How many numbers divisible by 5 can be made with the digits 2, 3, 4 or 5 if each digit is not used more than once?"

using all digits 3! => 6
using 3 digits 3!x2! =>12
using 2 digits 3
using 1 digit 1
total = 22

i think my answers are correct, not sure.

"Also i just want to clarify a answer which i didn't get to one question but i fill i am correct!!!

Question is "in how many ways can the lettters of the word READER be arranged?"

For this question i did this as

6!
-----
2! x 2!

the 2! x 2! Is because of the two times that "r" is used and "e"!!! and i get an answer of 180, however in the answers it says its 360????? can anyone confirm my thoughts on the answers being wrong or if its just me"

that answer is correct

 

[Kutay]

UE NMBR

This means 5 occupies the last digit. So basically it's how many arrangements of 2, 3, and 4 there are when the last digit has already been decided. So it's 3! = 6.

.

The answer u got of 3! = 6, the book says the answers 16!!!! Are you sure u haven't made a mistake, but by what u have said it makes sense or has the book just made another mistake or something ?

 

[Kutay]

"In a random arrangement of the letters of the word NUMBER what is the probabilty that the two vowels will come together?"

total arrangements = 6! => 720
vowels together arrangements = 5!x2! => 240
probability 360/720 => 1/3

"How many numbers divisible by 5 can be made with the digits 2, 3, 4 or 5 if each digit is not used more than once?"

using all digits 3! => 6
using 3 digits 3!x2! =>12
using 2 digits 3
using 1 digit 1
total = 22

i think my answers are correct, not sure.

"Also i just want to clarify a answer which i didn't get to one question but i fill i am correct!!!

Question is "in how many ways can the lettters of the word READER be arranged?"

For this question i did this as

6!
-----
2! x 2!

the 2! x 2! Is because of the two times that "r" is used and "e"!!! and i get an answer of 180, however in the answers it says its 360????? can anyone confirm my thoughts on the answers being wrong or if its just me"

that answer is correct

MY ANSWER???? or the book's answer?

 

[Dreamerish*~]

The answer u got of 3! = 6, the book says the answers 16!!!! Are you sure u haven't made a mistake, but by what u have said it makes sense or has the book just made another mistake or something ?

Oh gosh! My bad, I was thinking they had to be 4-digit, but they dont.

4-digit: ³P₃ = 6
3-digit: ³P₂ = 6
2-digit: ³P₁ = 3
1-digit: It can only be 5 so = 1

6 + 6 + 3 + 1 = 16.

There it is.