Combinatorics

bobbym · 2013-09-21 05:45:29

That makes sure there are 4 different types and removes duplicates.

anonimnystefy · 2013-09-21 05:48:26

But, there aren't 4 types of pens.

bobbym · 2013-09-21 05:50:52

Yes there is, type 1,2,3, and 4. They are the second number in each list.

anonimnystefy · 2013-09-21 05:53:05

Nowhere in the problem is stated that we can't choose both red1 and green1. That is the main problem with the definition of ans2.

anonimnystefy · 2013-09-21 05:54:50

Here is a code that gets the correct answer

s = {{r, 1}, {r, 2}, {b, 1}, {b, 2}, {b, 3}, {g, 1}, {g, 2}, {g, 3}, {g, 4}};
ans = Permutations[s, {4}];
ans1 = Select[ans, Length[Union[#[[All, 1]]]] == 3 &];
ans2 = Union[Sort /@ ans1] // Length

Last edited by anonimnystefy (2013-09-21 05:55:02)

bobbym · 2013-09-21 05:58:46

Okay, let me try that.

anonimnystefy · 2013-09-21 05:59:35

I have noticed it is pretty much the same as yours except it doesn't have the ans2 line.

bobbym · 2013-09-21 06:01:50

How does it deal with this part?

You want to purchase 4 pens, each of a different type

Where is that condition done?

anonimnystefy · 2013-09-21 06:05:08

It's the ans1 line, which is the same as yours.

bobbym · 2013-09-21 06:07:13

The first correct answer is

But if the second numbers are the types, this one only has 2 different types, not 4.

anonimnystefy · 2013-09-21 06:08:51

Those numbers are used just for differentiating the pens.

bobbym · 2013-09-21 06:14:53

I think you are correct but I do not agree with the problem wording. It is my major dislike of that site.

anonimnystefy · 2013-09-21 06:17:04

If the problem with this problem's wording is that it says type, you should know type refers to the colour.

What the problem with that site is, is most of the problems are user submitted and probably not very much checked.

bobbym · 2013-09-21 06:19:38

That is where I did not get that idea and came up with a different answer.
Luckily this was not in my problem set.

Bob · 2013-09-21 11:45:42

I have a complete list. I used my method of post 2, eliminating repeats. There are 72.

Bob

anonimnystefy · 2013-09-21 11:47:37

144?

Bob · 2013-09-21 11:52:04

No 72. I just re-read the question and took out the repeated types.

It's just 24 x 6 divided by 2.

Bob

anonimnystefy · 2013-09-21 12:00:01

Ah, okay.

Bob · 2013-09-21 14:32:21

Here they are:

Bob

Agnishom · 2013-09-21 15:14:35

I don't really understand how GFs are used. Could anyone teach me?

bobbym · 2013-09-21 20:51:35

You asked that question in the other forum that we are both members of. I believe it was about partitions.

Generating functions are a way to change a combinatorics problem into a computational one. What would you like to do with them?

Agnishom · 2013-09-21 22:55:15

this problem.

stefy's forum? I've lost the link

bobbym · 2013-09-21 22:59:02

http://maths.freeforums.org/

Did you expand his gf?

We also had a good discussion about gf's right here.

http://www.mathisfunforum.com/viewtopic … 87#p268687

Agnishom · 2013-09-22 02:07:01

We had, but I'm not confident about it

anonimnystefy · 2013-09-22 02:16:43

Hi Agnishom

You can also check out this thread: http://www.mathisfunforum.com/viewtopic … =17704&p=2. Bobbym gave some interesting examples in that thread which use GFs like I did for the problem in this thread.

Math Is Fun Forum

#26 2013-09-21 05:45:29

Re: Combinatorics

#27 2013-09-21 05:48:26

Re: Combinatorics

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Re: Combinatorics

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Re: Combinatorics

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Re: Combinatorics

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Re: Combinatorics

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