Permutations help (1 Viewer)

[clintmyster]

Someone help me with these questions please..really stumping me..thanks!

1. The first ten letters of the alphaet are to be placed in a line, find the number of arrangements possible if:

a) the A and B are together and the C, D and F are not.
b) The A and B are together, the C and D are together and the H and J are not together

2. Eight people, David, Anne, Mike, Mandy, Sue, Adam, John and Emma, enter a picture theatre and decide to sit in the back row. How many different seating arrangements are there if:

a) David and Anne are to sit together and John and Sue are not to sit together

 

[lacklustre]

Shouldn't this be in the Extension 1 forum?

 

[foram]

1. The first ten letters of the alphaet are to be placed in a line, find the number of arrangements possible if:

a) the A and B are together and the C, D and F are not.

9!x2!-7!x3!

b) The A and B are together, the C and D are together and the H and J are not together

8!x2!x2!-7!x2!

2. Eight people, David, Anne, Mike, Mandy, Sue, Adam, John and Emma, enter a picture theatre and decide to sit in the back row. How many different seating arrangements are there if:

a) David and Anne are to sit together and John and Sue are not to sit together.

7!x2!-6!x2!

I don't know if i'm right or not... i've never done these types before.

 

[lolokay]

I would have thought it would be
1a) 9!2! - 7!3!2!

1b) 8!2!2! - 7!2!2!2!

2) 7!2! - 6!2!2!

 

[clintmyster]

its extn? i wasnt sure so i just posted here..as for the answers..what ive been given is

1a) 302400
b) 120960

2a) 7200

just need the working to understand how to get there!

 

[foram]

this is harder than anything thats going to show up on your test.

 

[clintmyster]

i hope it doesnt..got a 3u test on wed and like they expect us to learn permutations and combinations by ourselves =[ its part of the yr12 3unit book as well!

 

[lolokay]

2 of my answers were correct.
You are finding the permutations of the single members + included groups, subtracting the permutations of single members + included groups + excluded groups.

For the first one, 1a - does it mean any 2 of C, D and F cannot be together? I assumed it meant only all 3 cannot be together. I'll figure out how to do it with that as a condition in a sec

EDIT:

1a) is 6!2!7!/4!

1b) is also 6!2!2!7!/5!

2. is also 5!2!6!/4!



These are because: let each thing and each pair of things each considered a group (things that have a condition that they cannot be next to another thing are not a group). Let m.k be number of things per group. The number of ways to order the groups, n, is n! * m.1! * m.2! .. m.n! (you only need to bother with groups > 1). Now, the number of ways to sort the things, x, that are not included, ie had a condition that they could not be next to another thing, is (n+1)!/(n+1-x)!, so multiply this by the number of ways to order the groups.

That probably was really hard to understand, and I don't know what the proper name for 'things' is


EDITEDIT:
Further explanation:
n is the number of 'groups'. m is the things per group. k is the group number (doesn't matter which group is numbered which; just imagine them in a row and assign each a number from 1 - n. Basically, it means that for every group of two, eg "A and B have to be together", you multiply by 2!).
x is the number of things that arent groups.
ways of sorting is (n+1)!/(n+1-x)! because, there are n-1 places in between the groups, + 2 on the ends, for the first to go = n + 1; substitute n+1 for n, and x for k in formula for permutations of k things from n, = n!/(n-k)!

 

[clintmyster]

2 of my answers were correct.
You are finding the permutations of the single members + included groups, subtracting the permutations of single members + included groups + excluded groups.

For the first one, 1a - does it mean any 2 of C, D and F cannot be together? I assumed it meant only all 3 cannot be together. I'll figure out how to do it with that as a condition in a sec

EDIT:

1a) is 6!2!7!/4!

could you expand on your working to help me out a bit? like why each factorial is the case?

 

[lolokay]

I added it to my post. let me know if it makes sense/if not then where

 

[clintmyster]

I added it to my post. let me know if it makes sense/if not then where

your whole explanation is a bit complex for me..can it be any more simpler?

 

[lolokay]

your whole explanation is a bit complex for me..can it be any more simpler?

lol, I thought it would be, it's pretty poorly written..

for 1a, firstly ignore the C,D,F. The number of ways to arrange the remaining 7 letters is 6!2! (as the A and B are together, so treated like one letter, but can be arranged 2! ways).

Now, there are 7 places where the C,D and F can go (in between each of ther other letters, and on the ends). Only one can go in each place as those letters cannot be next to each other.

The way of arranging the C, D and F is therefore 7*6*5 = 7!/4!

So there are 6!2!7!/4! ways altogether

EDIT: got the numbers wrong