Three Permutations and Combinations Questions (1 Viewer)

[Drsoccerball]

Braintic liked my post which means he agreed with me... This book didnt get any of those questions right what book is this?

In fact the combinations of no repeats is 210 which is greater than your textbook answer....

 

[braintic]

Would you please explain your reasoning for this?

As I said below, if there is a group of k, then the other n-k also form a group.
There isn't any freedom to arrange two groups around a circle - there is only one way.
The k! and (n-k)! come from arranging within the groups.

And it looks like your textbook got all three answers wrong.
Which textbook is this?

 

[Ambility]

Braintic liked my post which means he agreed with me... This book didnt get any of those questions right what book is this?

You're going to love this... MIF. The only reason I'm using it is because Cambridge doesn't seem to have a P&C section.

 

[Drsoccerball]

As I said below, if there is a group of k, then the other n-k also form a group.
There isn't any freedom to arrange two groups around a circle - there is only one way.
The k! and (n-k)! come from arranging within the groups.

And it looks like your textbook got all three answers wrong.
Which textbook is this?

Nek minute carrot sticks new book...

 

[Drsoccerball]

You're going to love this... MIF. The only reason I'm using it is because Cambridge doesn't seem to have a P&C section.

Okay now i know why people say the book sucks ass

 

[Ambility]

The (n-k)! is for arranging the two groups. The k! arranges the people within the group containing k people.

Why wouldn't the first part be (n-k+1)! ?

The number of "things" being arranged is n-k (the amount of people not in the group) and one more, the group itself. For example, if you have a 5 people with 3 of those people in a group, you have the two people not in the group, and the group itself. That makes 3 things. You are suggesting there are only two things to be arranged.

EDIT: You know what, I sort of get it with your further explanation of one group being (n-k)! and the other being k!.

 

[braintic]

Why wouldn't the first part be (n-k+1)! ?

The number of "things" being arranged is n-k (the amount of people not in the group) and one more, the group itself. For example, if you have a 5 people with 3 of those people in a group, you have the two people not in the group, and the group itself. That makes 3 things. You are suggesting there are only two things to be arranged.

His explanation is not right. There is only one way of arranging the two groups.

But doing it the way you are describing, the number of ways of arranging (n-k+1) things in a circle is (n-k)!
(You have learned this rule?)

 

[Ambility]

I am so fucking done with this textbook.

 

[braintic]

I am so fucking done with this textbook.

Perhaps it should be called Astigmatic Maths.

 

[Ambility]

His explanation is not right. There is only one way of arranging the two groups.

But doing it the way you are describing, the number of ways of arranging (n-k+1) things in a circle is (n-k)!
(You have learned this rule?)

Yeah, I've learnt that rule. This makes sense now.

 

[InteGrand]

The Year 12 3 Unit Pender (Cambridge) textbook has a chapter on perms and combs, so use that. Whatever you do, do NOT rely on MIF.

 

[Ambility]

The Year 12 3 Unit Pender (Cambridge) textbook has a chapter on perms and combs, so use that. Whatever you do, do NOT rely on MIF.

Lucky I bought that one after yesterday!

 

[pzeait]

for 1) it is 4! x 3! / 2! = 72. The first 3!/2! is the number of ways of arranging 3, 3 and 7.

 

[braintic]

for 1) it is 4! x 3! / 2! = 72. The first 3!/2! is the number of ways of arranging 3, 3 and 7.

As I said in an earlier post, the answer cannot possibly be as high as 72.