MedVision ad

MX1 Topic - Permutation questions (1 Viewer)

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
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.
 
Last edited:

deterministic

Member
Joined
Jul 23, 2010
Messages
423
Gender
Male
HSC
2009
(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

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
Wow thankyou heaps! Exactly the description I was looking for.
 

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
One more question added.\
Another one added.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top