Urgent: Maths CIE Stats

[$!$RatJumper$!$]

I can't really explain the workings from the marking scheme but I did it another way which was approved by my teacher.

First we find out the number of arrangements in which all the 9 persons can be arranged. This will be (9! +(7! x 3!))
9! => There are nine persons which can be arranged in any way.
(7! x 3!) => There are 6 men and 3 women. We'll be taking all the women to be standing beside each other. Hence they are taken as one possiblity only which makes 7!. The 3 women can then be arranged in any way 3!.

Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
8! = > One woman is taken as one possiblity and the other two as another possiblity. In all there will be 8 possiblities, after adding the six men.


Number of arrangements : (9! +(7! x 3!)) - 3(8! x 2!) = 151200

Hope you understand

Good job!

[Deadly_king]

Good job!

Glad to have been of help

I just hope cs understands it as well

[cs]

yea. there's working behind my pass year book, and they gave 2 types of working, i did the way you did, and manage to get 9! and 7!X3!, i don't get this part(when two women are together):

Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)

I got 8! X 2!, why is there a 3?

[Deadly_king]

yea. there's working behind my pass year book, and they gave 2 types of working, i did the way you did, and manage to get 9! and 7!X3!, i don't get this part(when two women are together):

Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)

I got 8! X 2!, why is there a 3?

The three is because there are three women

Let there be 3 women : W₁, W₂ and W₃

Possible arrangement are :
i) W₁W₂ and W₃
ii) W₁W₃ and W₂
iii) W₂W₃ and W₁

So there are 3 ways in which the women could be arranged

[S.M.A.T]

DK son ............u r awesome .................+rep

[Deadly_king]

DK son ............u r awesome .................+rep

Thanks daddy

Son resembles his dad

[cs]

Ohh i always thought its 3!, there's only 3 ways, ok i understand now.. Thank you!

[Deadly_king]

Ohh i always thought its 3!, there's only 3 ways, ok i understand now.. Thank you!

Glad you understood

[$!$RatJumper$!$]

Question guys

There are 3 married couples and 4 single people

What are the ways of choosing 5 people from these if:
i) 2 Married couples are chosen
ii) 1 Married couple is chosen

My answers i) 3C2 * 6C1 = 18
                ii) 3C1 * 8C3 = 168

[Deadly_king]

Question guys

There are 3 married couples and 4 single people

What are the ways of choosing 5 people from these if:
i) 2 Married couples are chosen
ii) 1 Married couple is chosen

My answers i) 3C2 * 6C1 = 18
                ii) 3C1 * 8C3 = 168

Do you mind saying which year it is?

i think you missed something in the question here

[$!$RatJumper$!$]

even i think i did miss something. this is a question my friend asked me that he got from a handout worksheet so its not from a paper. anyways dont worry about t bro

[Deadly_king]

even i think i did miss something. this is a question my friend asked me that he got from a handout worksheet so its not from a paper. anyways dont worry about t bro

Ok........if you are saying so.

But I've done a number similar to this, however i can't remember which year

[$!$RatJumper$!$]

so which exams do you have next buddy?

[Deadly_king]

so which exams do you have next buddy?

Statistics (S1) and two chem papers on wednesday. I believe it's the same for you, nan?

Then physics practicals on thursday

[$!$RatJumper$!$]

yes thats right thought i have computing tomorrow :/ haha