Melody

grs75.

I owe you an appology.

The question that you are now asking about, that Hectictar answered, was not your question.

So there was no reason why you ever needed to respond to it.  

Maybe you only just saw it. 

I had it confused with this one that Max answered that you did not respond to.

Thanks Grs75.

Your response is appreciated.

It seems my part A and your given answer are the same anyway.

That is a good start :)

We will expect a response from you grs75.

Preferably with the answer you are given as correct.

So far you do not have flawless history of showing appreciation.

Hi Hajer   

I do not want to do it all for you.

Start by finding the prime factors of 1200.   

You should know immediately from that what 2 of the rolls will have to be.

There are only 4 possibililites for what the rest will be.

then you just have to work out how many ways each of those 4 sets of 6 numbers can be ordered.

Give it a go. :)

nice logic builderboi but you did make a small error.

You can fix it if you want but you do not need to fix it, 

Guest can find it for him/her self.   

Here is my take.

A

In how many ways can five red chairs and three white chairs be arranged such that no two white chairs are next to each other?

I am also assumiong that all the white chairs are identical and all the red chairs are identical.

And I am assuming that the chairs are placed in a row

The white chairs must be seperated so I can start with 

W R W R W

Now there are 3 more red ones so I will put hashes  where one or more of them can go

# W # R W # R W #

If all three are together then there are 4 places they can go

If 2 are together and the other seperate then there are 3+3+3+3=12 ways

If none are placed together then we have  4C3=4ways

4+12+4= 20 ways

20 ways is my answer

B

B. In how many ways can four red chairs, three white chairs, and two blue chairs be arranged such that there is at least one red and one blue chair between two white chairs?

The white chairs are seperated by at lease 1 red and one blue chair so there is the start

W [BR] W [BR] W   

Of course the BR  could also be RB and that goes for the second lot of BR as well so there are 4 copies of all continuing arrangements

(I could be double overcounting but I don't think so)

So

W [BR] W [BR] W      Ive used all the White and all the blue chairs already there are only 2 red ones unused so where can I put them?

# W #[BR]# W #[BR]# W #  

If they both stay together then there are  6 places they can go

If they are seperated then there are  6C2 = 15 places they can go.

Which adds to 15+6=21 places

But there were 3 other starter arrangments 

4*21 = 84 possible arrangments