permutation queston

[Mortal9090]

(h) You are arranging six of your friends Alice, Bob, Charles, Diana, Francine, and George, in a row so that you can take their picture.

(i). Alice and Bob have had a fight and refuse to stand next to eachother. How many ways are there to arrange your six friends?

 

[UAEgurl]

24?

 

[ahmed t]

well for the first on its 6! simple
for the second
find how many ways that alice and bob are together
that would be 5! since we take alice and bob as one person and times by 2! since you can interchange alice and bob
so ur anser will be
(h)-2!5!

 

[IgoforA]

h) 6! and i) 5! *4! i guessss... haaah

 

[ahmed t]

so it will 480 for the second one

 

[Mortal9090]

i) = 2 x 6! - 2x2x5! = 960

i dont know how

 

[IgoforA]

Really? where do you get these questions? HERO... haha

 

[ahmed t]

its wrong then
im sure of my anser

 

[Mortal9090]

2x5! + 2x5! = 2x2x5! is the right answer,

i dont know why

 

[ahmed t]

2x2x5! which is 480
same as my anser.
i do not know how they have approched this question but i am sure my way of solving is correct.

 

[crash250]

the second amswer is 480

 

[ahmed t]

exactly^^^^^^^^^^

 

[IgoforA]

http://www.xtremepapers.me/CIE/Internat ... _qp_62.pdf <- similar like this paper..... for Q4 b)....

 

[memyself]

6 friends minus the two who refuse to stand next to each other, 5P2 x 4! = 480.
5P2 (2 out of the 5 spaces beside 4 friends) and 4! (the four friends left)

 

[djdead1]

isnt it simpler dis way? so many of u r misreadin da question =S

ways to arrange them, when 2 people out of 6 DONOT want to b 2geder
total ways --- ways in which both are ALWAYS together

= 6! --- (4! * 2!)
= 672

this givs the ways in which they cn b aranged whn the 2 DONT wana b 2geder.

(-- stands for minus)


plz teme what year dat question is frm, and is dis da ryt answer?

 

[ahmed t]

its not
the right anser is 480

 

[stxjunk]

(h) You are arranging six of your friends Alice, Bob, Charles, Diana, Francine, and George, in a row so that you can take their picture.

(i). Alice and Bob have had a fight and refuse to stand next to eachother. How many ways are there to arrange your six friends?

For the first part, the answer is simply 6!=720 ways
For the second part, you work out as follows:
number of ways in which alice and bob may be together= 5!x2!=240
number of ways in which they wont be together= 720-240=480 ways

thats simple as that

 

[Amberkgkl]

4! * 5P2
ans is 480.

 

[djdead1]

ouch ! i see da mistake i ws makin...thnks ppl

 

[reekx94]

Answer is 480. full stop.


here's how its done, and any question that says 2 people are not to stand next to each other.

if we ignore those 2 guys for a while, we are left with 4 guys to arrange. which is simply 4!

1_2_3_4_5

the number represent spaces, and the lines represent those 4 people to arrange. now we said there are 4! ways to arrange those right?
what about the two other people that had a fight. well Alice can stand in space 1, or 2, or 3, or 4, or 5. since she has to be seperated by one of the other 4 so that she doesnt stand next to bob.

well so alice can stand at 5 different places, then bob has 4 different places left since alice is gona take one of them. and therefore the answer is 4! x 5 x 4.

and by the way that is a very standard question, and therefore you dont have to think when it comes. a simple standard way is......

If you have N group of people, and 2 people don't wanna stand next to each other, then the different arrangements are..

(N-2)! x (N+1) x N. try it in different questions and check if it works, i invented this rule

Oh and if its 3 people who dont wanna stand next to each other then its (N-3)! x (N+1) x (N) x (N-2).......if its 4 you just continue (N-3)....

hope this helps