How to calculate sets of arrengements, who is right?

[Taniuca]

Hello,
I'm confused and I don't know who is right.

Problem:
six children A,B,C,D,E,F are to be sitted together in a single row of six chairs. If A cannot sit next to B, how many different arrangements of six children are possible?

a) 240, b) 480, c) 540, d) 720, e) 840.

From what I read is that I should take 2 of the children as if it was 1, so then I will have 5! posibilities = 120
Then I take the 2 that are sitting together and find the possible sits for those 2; that equals to 2!=2
I multiply 120*2= 240 options.

Why in this problem do I have to substract 6!- 240 to equal 480 which is the right answer while in the following problem I don't have to?

Problem:
find the arrangements of 8 people if 3 must be sit together.
correct answer= 4320

I take 3 as a 1 person so I find 6!= 720, then I take the 3 that are sitting together and find the possible sits in this case 3!= 6
Then I multiply 720*6= 4320

[goyalsau]

Hello,
I'm confused and I don't know who is right.

Problem:
six children A,B,C,D,E,F are to be sitted together in a single row of six chairs. If A cannot sit next to B, how many different arrangements of six children are possible?

a) 240, b) 480, c) 540, d) 720, e) 840.

From what I read is that I should take 2 of the children as if it was 1, so then I will have 5! posibilities = 120
Then I take the 2 that are sitting together and find the possible sits for those 2; that equals to 2!=2
I multiply 120*2= 240 options.

Why in this problem do I have to substract 6!- 240 to equal 480 which is the right answer while in the following problem I don't have to?

Problem:
find the arrangements of 8 people if 3 must be sit together.
correct answer= 4320

I take 3 as a 1 person so I find 6!= 720, then I take the 3 that are sitting together and find the possible sits in this case 3!= 6
Then I multiply 720*6= 4320

In first question they are asking they can not sit together, and in second they are sitting together...........

[Rahul@gurome]

In the problem of 6 children A, B, C, D, E, F the question is asking that A and B cannot sit next to each other. However in the second problem, 3 people have to sit together.

So for the first problem, number of ways of arranging A, B, C, D, E and F is 6!.
If A and B are together, then total number of arrangements possible is (6-2+1)!*2 or 240.
So the number of arrangements when they are not together is 6! - 240 or 480.