Perms and Comb Q's (1 Viewer)

[cutemouse]

1) There are 12 videotapes arranged ina row on a shelf in a video shop. THere are 3 identical copies of 'Gone in with the wind', 4 of 'Tootsie' and 5 of 'Gladiator'.

(i) How many different arrangements are there?

(ii) How manu different arragements are there if the videos with the same titles are goruped togehter?

(iii) The 12 videotapes are arranged at random in a row on the shelf. Find the probability that the arragement has a copy of 'Gone with the wind', at one end of the row, and a copy of 'Galdiator' at the other end?



2) Ten people arrive to eat at a restuarant. The only seating avilable for them in two circular tables, one that seats six persons, and another that seats four. Using these tables, how many different seating arragements are there for the ten people?



3) In how many ways can 7 people sit at a round table so that 2 particular people:

(i) sit next to each other?

(ii)are seperated?



Thanks

 

[ClockworkSoldier]

Loooooooooool.

Me being the audio nerd that I am, saw "Comb Q" and instantly thought that you meant the Q on a parametric EQ .

Just thought I'd tells youse .

 

[Timothy.Siu]

1) There are 12 videotapes arranged ina row on a shelf in a video shop. THere are 3 identical copies of 'Gone in with the wind', 4 of 'Tootsie' and 5 of 'Gladiator'.

(i) How many different arrangements are there?

1i)12!/3!4!5!

(ii) How manu different arragements are there if the videos with the same titles are goruped togehter?

3! (since theres 3 groups of things, and theres only one way of arranging each group)

(iii) The 12 videotapes are arranged at random in a row on the shelf. Find the probability that the arragement has a copy of 'Gone with the wind', at one end of the row, and a copy of 'Galdiator' at the other end?

well, lets take out one gone with the wind and one gladiator
now we have 10 objects.
arranging them, 10!/2!4!4! but they can we at either end. so,
(2x10!/2!4!4!)/(12!/3!4!5!)=5/22

2) Ten people arrive to eat at a restuarant. The only seating avilable for them in two circular tables, one that seats six persons, and another that seats four. Using these tables, how many different seating arragements are there for the ten people?

seat one table at a time, 10C6 x 6!/6 x 4C4 x 4!/4

3) In how many ways can 7 people sit at a round table so that 2 particular people:

(i) sit next to each other?

we can make those 2 people into one group. theres 2 ways of arranging that group.
so, now we have to arrange 6 objects at a round table.
6!/6 x 2 is the answer=240

(ii)are seperated?

lets arrange everyone else first, so, 5!/5 is 5 people.
then let's slot the 2 people in between other people so they cant be together.
theres 5 slots. so, 5!/5 x 5P2=480

 

[youngminii]

Alternative method to figure out some of the answers (timsiu is much smarter than me so you'd probably want to use his methods this is just for completeness)
1. i) 12C5 x 7C4 x 3C3 = 27720
iii) 3/12 (chances of a gone with the wind being on one end) x 5/11 (chances of a gladiator being on the other) x 2 (they can be on either side) = 5/22
OR
If we take out one gone with the wind and one gladiator then there are
10C4 x 6C4 x 2C2 x 2 (they can be on either side) = 6300 ways of arranging the books to have a gladiator and a gone with the wind on opposing ends. Now divide that by the total number -> 6300/27720 = 5/22

3. ii) This is the exact opposite of i), so we can do 7!/7 (total number of arrangements of the table) minus answer of i) = 6! - 5! x 2 = 480
I don't know how to do perms like tim did in the last question

 

[cutemouse]

(iii)

well, lets take out one gone with the wind and one gladiator
now we have 10 objects.
arranging them, 10!/2!4!4! but they can we at either end. so,
(2x10!/2!4!4!)/(12!/3!4!5!)=5/22

Hi, I did (3C2 * 5C2)/(12C2) for that, which is 5/11. Is that somewhat correct, or completely wrong?

Thanks

 

[cutemouse]

seat one table at a time, 10C6 x 6!/6 x 4C4 x 4!/4

Where'd you get the 10C6 and 4C4 from?

 

[Timothy.Siu]

Where'd you get the 10C6 and 4C4 from?

i'm choosing people to sit at the table then arranging them

 

[Timothy.Siu]

(iii)


Hi, I did (3C2 * 5C2)/(12C2) for that, which is 5/11. Is that somewhat correct, or completely wrong?

Thanks

uhh, i'm not really sure what you're doing.
i sorta know how ur trying to do it.
maybe do,
5/12 for one of them to be gladiator
then, 3/11 for the other one. but they can be either end.
so, 5/12 x 3/11 + 3/12 x 5/11=5/22

 

[cutemouse]

i'm choosing people to sit at the table then arranging them

Thanks for your replies.

Hmm, if I understand correctly you're arranging 10C6 people for the table that can seat 6 people, so the remaining people are 4 people, which can be seated in 4C4 ways.

Sorry to sound so dumb, but why do we need to do that? Why can't we just do 5!*4! ?

Thanks again.

 

[Timothy.Siu]

theres more ways to arrange 10 people at 6 seats than 6 people at 6 seats

 

[cutemouse]

Ahh okay... That kinda explains it.

Was what I said before correct?

ie. that there are 10C6 ways to arrange 10 people in 6 ways in 5! ways... and that the remaining 4 people can be arranged in 4C4=1 way in 3! ways.

Thanks

 

[Timothy.Siu]

yeah, if what i think you're saying is right

 

[cutemouse]

Alrighty thanks...

I have another question on this topic... -_-;

Don guess, at random, the answers to each of 6 multiple choice questions. IN each question there are 3 alternative answers, only one of which is correct.

(i) Find the probability that Don answers exactly two questions correctly (can do this one)

(ii) Find the probability that the 6th question that Don attempts is only the 2nd question that he answers correctly.

Thanks

 

[Timothy.Siu]

Alrighty thanks...

I have another question on this topic... -_-;

Don guess, at random, the answers to each of 6 multiple choice questions. IN each question there are 3 alternative answers, only one of which is correct.

(i) Find the probability that Don answers exactly two questions correctly (can do this one)

(ii) Find the probability that the 6th question that Don attempts is only the 2nd question that he answers correctly.

Thanks

i think u can just find the probability that he gets 1 right in 5 questions, then multiply that by 1/3