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MX1 Topic - Permutation questions (1 Viewer)

Skeptyks

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The question below is solved. I have another question, at a dinner party 6 people sit around a table.
b) Find the probability that if seating is at random, 2 friends will sit apart.

Also, how many arrangements are possible from the word STUDIO if the letter s is not the first letter. I did 5! x 5 and it gives the correct answer however I am not too sure why it is. Is it because the first letter can be any of the 5 other letters and afterwards, there is a choice of 5 letters, then 4 letters, then 3 letters etc?


I'm just starting to grasp the concept of permutations and combinations and I need help with this permutation question.

"A queue has 4 boys and 4 girls standing in a line. Find how many different arrangements of the line are possible if

The boys and girls alternate.
2 particular girls wish to stand together
all the boys stand together
also find the probability that 3 particular people will be in the queue together if the queue forms randomly"

I've been stuck from the first one, tried 7P4 x 6P3 but it just isn't working out. I have no idea how to approach this question.

Any help will be appreciated.
 
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deterministic

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(1) Boy girls alternate
It will either be
BGBGBGBG or GBGBGBGB
Arranging the boys in either case will be 4!. Same with the girls. So number of arrangements = 4!*4! (number of arrangements for each case) * 2 (number of cases)

(2) 2 girls together:
Treat 2 girls ("GG") as one group. Then you are arranging 4 boys, 2 individual girls and 1 couple (the "GG"). Then number of arrangements = 7!(arranging all items)*2! (number of arrangements within the group of girls)

(3) All boys stand together: Similar to above case, have a go at it yourself.

(4) Probability = number of favourable arrangements/number of total arrangements
So find the number of favourable arrangements (ie. 3 people together- similar to (2) and (3)) and find the total number of arrangements of 8 people (8!). Have a go at this yourself.
 

Skeptyks

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Wow thankyou heaps! Exactly the description I was looking for.
 

Skeptyks

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HSC
2012
One more question added.\
Another one added.
 
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