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Permutation Questions Help (1 Viewer)

csi

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Hi,

I really don’t get permutation, please help with the following:

1. The letters of the name JESSICA are arranged in a line, what is the probability that the S’s are at the ends?

2. In how many ways can four different maths books, two different physics books and five different english books be arranged on a shelf if books of the same subject are kept together?

3. From a pack of 9 cards numbered 1-9, three cards are drawn at random and laid on a table from left to right.
(a) what is the probability that the number formed exceeds 400?
(b) what is the probability that the digits are drawn in descending order?

4. There are 720 ways seven children can join hands to form a circle if the all face inward. In how many of these cases does child A joins hands with children B and C?

5. Three Tasmanian, three New Zealander and three people from NSW are seated at random around a round table. What is the probability that the people in the three groups are seated together?

THANK YOU!!
 

beetree1

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its been over a year since i touched on perms and coms so imma just do the ones i can be bothered to do:

2. you have 3 subjects so 3! then there's 4! x 2! x 5! for each subject arranged in their groups
so the answer should be 3! x 4! x 2! x 5! = 34560? could be wrong

5. same concept^ i seem to only know how to do these atm lol
(3-1)! x 3! x 3! x 3! = 432
 

TheShy

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Ill add on:
1. Total is 7!/2! cos of the repeating S. Then the number of ways where the S's are at the end : 5! x 2!/2!. So the probablity, if I didn't do any mistakes should be (5! x 2!/2!)/(7!/2!)

4. Think the question is trying to make itself seem tricky by adding the 720 ways and if they all face inwards. Think of it as a regular circle question. You got 7 people, and person 1 must sit next to person 2 and 3. There is only 2! ways of this happening. 4 people remaining, so they can be arranged 4!. So answer should be 2! x 4!. You'll have to check the answer on this, not sure if what I did was right.
 

tito981

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1. fix the position of the S's at the ends, so its 5!x2!, the 2! is for the swap between the S's.
3. (a) to be greater than 400 the first number that has to be chosen can't be 1,2,3 so there are 6 possibilities for the first one and then there is 9-1 then 9-2. mulitply all those together for the answer.
(b) you can do this by case, so if 9 is chosen all the other numbers can be chosen..., and do it for all the numbers.
4. child A has to be in between B and C, so just fix their positions and solve. (forgot exact process after)
 

csi

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its been over a year since i touched on perms and coms so imma just do the ones i can be bothered to do:

2. you have 3 subjects so 3! then there's 4! x 2! x 5! for each subject arranged in their groups
so the answer should be 3! x 4! x 2! x 5! = 34560? could be wrong

5. same concept^ i seem to only know how to do these atm lol
(3-1)! x 3! x 3! x 3! = 432
ty:)
 
Last edited:

csi

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Ill add on:
1. Total is 7!/2! cos of the repeating S. Then the number of ways where the S's are at the end : 5! x 2!/2!. So the probablity, if I didn't do any mistakes should be (5! x 2!/2!)/(7!/2!)

4. Think the question is trying to make itself seem tricky by adding the 720 ways and if they all face inwards. Think of it as a regular circle question. You got 7 people, and person 1 must sit next to person 2 and 3. There is only 2! ways of this happening. 4 people remaining, so they can be arranged 4!. So answer should be 2! x 4!. You'll have to check the answer on this, not sure if what I did was right.
much appreciated :)
 
Last edited:

csi

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1. fix the position of the S's at the ends, so its 5!x2!, the 2! is for the swap between the S's.
3. (a) to be greater than 400 the first number that has to be chosen can't be 1,2,3 so there are 6 possibilities for the first one and then there is 9-1 then 9-2. mulitply all those together for the answer.
(b) you can do this by case, so if 9 is chosen all the other numbers can be chosen..., and do it for all the numbers.
4. child A has to be in between B and C, so just fix their positions and solve. (forgot exact process after)
thanks:)
 
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