Math with Probability ?

[Dencology]

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Hey,
can anyone help me with this problem?

five people were to be arragnged in a row. how many ways can it be possible if two of them cannot be separated from one another ?

ans. 48.

i thought i would do this:

ABCDE and lets say that AB can not be separated from one another so we can do 4! = 4*3*2*1. and i t would be 24, but the ans is 48. i guess they multiply by 2 at the end. but why?

 

[polarmolar]

Hey,
can anyone help me with this problem?

five people were to be arragnged in a row. how many ways can it be possible if two of them cannot be separated from one another ?

ans. 48.

i thought i would do this:

ABCDE and lets say that AB can not be separated from one another so we can do 4! = 4*3*2*1. and i t would be 24, but the ans is 48. i guess they multiply by 2 at the end. but why?

I think the way you did it find all the possibilities if "A" and "B" cannot be seperated with "A" before "B", [AB] as one unit. It would be twice that because you can also have the same number of ways with "B" before "A", [BA] as one unit.

I think this is correct but ask around

 

[Dencology]

no it makes sense for what you have here. just like you said it could be AB or BA.

 

[klutzy1987]

You need to multiply 4! by the total different number of ways that that can be arranged. Since there are 2 different ways that it can be arranged as polar said AB or BA. So 24*2=48.

 

[Dmitry Malayev]

5 People, but 2 of them are always together, assume it's 1 person.

Therefore you have 4! = 4*3*2*1 = 24.

Now since the 2 people can be arranged in 2! = 2*1 = 2 you need to take that into account.

Hence 4!*2! = 24*2 = 48