Permutation (1 Viewer)

[mattchan]

Hey, i need some help with this question. I cant seem to get the 2nd part of this question. Answers are 2880 and 864

In how many ways can 5 men and 5 women be arranged in a cirlce so that the men are seperated? In how many ways can this be done if two particular women must not be next to a particular man?

Cheers

 

[Pace_T]

In how many ways can 5 men and 5 women be arranged in a cirlce so that the men are seperated? In how many ways can this be done if two particular women must not be next to a particular man?


men are separated.. ok so u got this:

men can be arranged 4! ways (1 less, because the first one doesn't count - its a circle)
.'. chicks have to be in 5! ways

.'. total arrangements 4!*5! = 2880

if two chicks cant be next to a guy then :

the two chicks have 3 choices to be seated. ie 3p2. the rest have 3p3

3p2 * 3p3 * 4! = 864

 

[mattchan]

Thanks dude. Heres another one lol

In how many ways can 8 girls be divided into 4 sets of 2. Answer 105

 

[Pace_T]

In how many ways can 8 girls be divided into 4 sets of 2. Answer 105

ok umm 8C2*6C2*4C2*2C2 all over 4!
= 105

 

[rama_v]

(8C2 x 6C2 x 4C2 x 2C2 )/4! which equals 105
damn, beaten to it! lol

 

[Pace_T]

Muahahaha

 

[rama_v]

How about some help with this one please guys:

In how many differnt orders can 10 examination papers be set so that no two of the three mathematics papers are consecutive?

answer is 42 x 8! (how do they get that?)

 

[Pace_T]

In how many differnt orders can 10 examination papers be set so that no two of the three mathematics papers are consecutive?

The papers can be X Y and Z
That means, X Y or Z can't be next to each other
You've got the other 7 papers (lets call them M)
M M M M M M M ( you lay out the 7 papers first because they don't have a condition like the X Y Z papers)
Now, the X Y Z can be anywhere in those spaces because they won't be next to each other (includes all the way on the left and on the right)
i.e. their are 8 spots to filled by 3 test papers
8P3
.'. arrangements = 8P3 * the ways the M's can be arranged
=8C3 * 7!
which is the same as ur answer.
This is in the Excel book. Is there any other ways of doing it?
Sorry if you still don't get it, I tried to make it simpler than their given solution

 

[rama_v]

In how many differnt orders can 10 examination papers be set so that no two of the three mathematics papers are consecutive?

The papers can be X Y and Z
That means, X Y or Z can't be next to each other
You've got the other 7 papers (lets call them M)
M M M M M M M ( you lay out the 7 papers first because they don't have a condition like the X Y Z papers)
Now, the X Y Z can be anywhere in those spaces because they won't be next to each other (includes all the way on the left and on the right)
i.e. their are 8 spots to filled by 3 test papers
8P3
.'. arrangements = 8P3 * the ways the M's can be arranged
=8C3 * 7!
which is the same as ur answer.
This is in the Excel book. Is there any other ways of doing it?
Sorry if you still don't get it, I tried to make it simpler than their given solution

Ahh I think i get it, so your basically saying that its 8P3 *7! because your counting two of the papers as one (since if two of them are not together than obviously 3 are not togther), is that right?

 

[currysauce]

i hate that friggen topic with a passion

 

[Pace_T]

Ahh I think i get it, so your basically saying that its 8P3 *7! because your counting two of the papers as one (since if two of them are not together than obviously 3 are not togther), is that right?

I think you are misinterpreting the question. It states that there are 10 mathematics papers. 3 of which are special ones that cannot be set next to each other. The other 7 are just ordinary exams.

There is not 3 different types of exams. They are just asking us to calculate the arrangements if 3 particular papers can't be next to each other. So we place the 7 remaining papers in a row. The 3 papers that can't be next to each other can be placed in any place between the 7 papers. With the 3 papers in 3 of these places, they won't be next to each other a there will be atleast 1 of 7 of the other papers between them.

 

[rama_v]

I think you are misinterpreting the question. It states that there are 10 mathematics papers. 3 of which are special ones that cannot be set next to each other. The other 7 are just ordinary exams.

There is not 3 different types of exams. They are just asking us to calculate the arrangements if 3 particular papers can't be next to each other. So we place the 7 remaining papers in a row. The 3 papers that can't be next to each other can be placed in any place between the 7 papers. With the 3 papers in 3 of these places, they won't be next to each other a there will be atleast 1 of 7 of the other papers between them.

Ahh now that explains it, I get it now. Thanks heaps