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Three Permutations and Combinations Questions (2 Viewers)

Ambility

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Hey everyone, please explain how to do these...

1. Numbers are formed from the digits 1, 2, 3, 3, and 7 at random. In how many ways can they be arranged to form a number greater than 30 000? (Answer is 72).

My thinking: In that question, I understand that the first number must be a 3, 3, or 7. After that number has been chosen, there are 4 other options for the next, 3 for the next, 2 for the next and 1 for the last. the double 3's can be swapped, so I'd divide by 2. I get 36.

2. A group of n people sit around a circular table. How many arrangements are possible if k people sit together? (Answer: (n-k+1)!/k!

My thinking: k people sit together, so I can consider that group of people as one person. The amount of people is now equal to n-k+1 (n-k for the people not in the group, and +1 for the group itself). Seeing as the first person (or group) can sit anywhere, the amount of arrangements of people should be equal to (n-k+1-1)! = (n-k)!.

3. How many different arrangements are possible if 3 letters are randomly selected from the word CHALLENGE and arranged into 'words'?

My thinking: I could do this if there weren't any double letters. How do I handle the double letters in the scenario? Some 3 letter words will have double letters, some wont.
 

DatAtarLyfe

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1/ So your right in saying that the first number has to be either a 3,3 or 7. So in the first slot out of the five, there are three options. Since there are no restrictions on the other four slots, the rest of the numbers can be arranged in any order i.e. 4!
Therefore, the answer is 3 x 4! = 72
 

Drsoccerball

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1/ So your right in saying that the first number has to be either a 3,3 or 7. So in the first slot out of the five, there are three options. Since there are no restrictions on the other four slots, the rest of the numbers can be arranged in any order i.e. 4!
Therefore, the answer is 3 x 4! = 72
But 73321 is the same as 73321 (3's switch)
therefore OP is right 4!3/2!
 

rand_althor

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3. How many different arrangements are possible if 3 letters are randomly selected from the word CHALLENGE and arranged into 'words'?

My thinking: I could do this if there weren't any double letters. How do I handle the double letters in the scenario? Some 3 letter words will have double letters, some wont.
You have to consider different cases. So one case could be 3 unique letters (e.g. CHA), and another a pair of identical letters and one unique letter (e.g. CLL).
 

braintic

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The number of numbers without restriction is 5!/2! = 60.

So the answer can't be 72.

The correct calculation is 4! + 4!/2!.
 

braintic

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The answer to 2 is also wrong.

It is just k! (n-k)!
 

braintic

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My bad its n-k since the total terms is n-k+1 (if bundell k) and since its a circle you subtract one therefore (n-k)!
It is easier than that.
If there is a group of k, then the other n-k also form a group.
 

Ambility

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The number of numbers without restriction is 5!/2! = 60.

So the answer can't be 72.

The correct calculation is 4! + 4!/2!.
That seems logical, the quantity of numbers about 30 000 should be less than the total quantity of numbers. I believe it should be , which is equal to 36, the same as your calculation. The textbook definitely give 72 as the correct answer, however, at this stage I guess it's safe to say we are right, and the textbook is wrong.
 

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