Combinatorics

Agnishom · 2013-09-20 13:16:14

There are 2 types of red pens, 3 types of blue pens, and 4 types of green pens.
You want to purchase 4 pens, each of a different type, containing at least one of each color.

In how many ways can you do this?

Bob · 2013-09-20 20:06:17

There are 2 types of red pens, 3 types of blue pens, and 4 types of green pens.
You want to purchase 4 pens, each of a different type, containing at least one of each color.

Let's say you have these colours red1, red2, blue1, blue2, blue3, green1, green2, green3, and green4. That's nine colour choices.

So choose a red, then a blue, then a green .... how many ways ?

Then choose anything as the fourth pen out of the 6 remaining choices.

Bob

anonimnystefy · 2013-09-20 21:32:34

Hi Bob

That is not going to yield a correct answer.

Hi Agnishom

Are pens of same color the same or different?

Bob · 2013-09-20 22:47:47

My first attempt at this question interpreted the problem differently. Then I deleted it and tried again. Now I'm not sure. We await Agnishom's clarification.

Bob

anonimnystefy · 2013-09-20 22:54:09

I'd say the answer is 3 if same-coloured pens are the same and 72 if they are different.

The GFs are:

and

Last edited by anonimnystefy (2013-09-21 00:18:29)

Agnishom · 2013-09-21 00:13:06

I think they are different

Agnishom · 2013-09-21 00:19:05

bob bundy wrote:

There are 2 types of red pens, 3 types of blue pens, and 4 types of green pens.
You want to purchase 4 pens, each of a different type, containing at least one of each color.
Let's say you have these colours red1, red2, blue1, blue2, blue3, green1, green2, green3, and green4. That's nine colour choices.
So choose a red, then a blue, then a green .... how many ways ?
Then choose anything as the fourth pen out of the 6 remaining choices.
Bob

So, it should be 2*3*4*6 = 144. But why is it 72?

anonimnystefy · 2013-09-21 00:24:07

I already said that isn't correct. If you did it like that, you would count picking red1, red2, blue1, green1 and red2, red1, blue1, green1 as different picks, when they are truly the same.

anonimnystefy · 2013-09-21 00:32:34

Hi Agnishom

I checked the answer with them and it is correct.

Agnishom · 2013-09-21 03:54:21

anonimnystefy wrote:

I already said that isn't correct. If you did it like that, you would count picking red1, red2, blue1, green1 and red2, red1, blue1, green1 as different picks, when they are truly the same.

I am sorry, I could not follow. How is it coming to 72?

I too checked that it is correct

anonimnystefy · 2013-09-21 04:06:37

You can do it by casework or using the GF above.

If you did it by casework you'd do it by choosing one colour and calculating the number of possibilities in which you buy two lens of that colour and one pen of each other colour. Then do that for the two other colour and sum them. The result should be: 1*3*4+2*3*4+2*3*6=72.

bobbym · 2013-09-21 04:09:04

Hmmm.

anonimnystefy · 2013-09-21 04:35:59

I don't know how I would program this one, and it's a simple enough a problem that it doesn't need to be programmed.

bobbym · 2013-09-21 04:40:45

Hmmm.

I don't know how I would program this one

So then it is not so simple. DZ says you do not understand the problem until you program it. This always lends insight and satisfies the "two solution rule."

anonimnystefy · 2013-09-21 04:42:25

I have two solutions. Classic casework and the GF.

bobbym · 2013-09-21 04:47:08

And if the problem were say 20 pens and 16 to 1 types would you still want to casework it?

anonimnystefy · 2013-09-21 04:53:49

bobbym wrote:

And if the problem were say 20 pens and 16 to 1 types

What?

bobbym · 2013-09-21 04:58:57

What I am saying is casework is a very clumsy way of working sufficient for small problems only.

You are right, it is difficult to program though.

anonimnystefy · 2013-09-21 05:12:39

I know, I dislike casework, too, but tend to use it if it seems possible.

Have you programmed it?

bobbym · 2013-09-21 05:17:50

Yes I did. But it refuses to get the answer I want.

anonimnystefy · 2013-09-21 05:20:54

Can you post the code you currently have?

bobbym · 2013-09-21 05:24:55

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 = Select[ans1, Length[Union[#[[All, 2]]]] == 4 &];
Union[Sort[#] & /@ ans2]

This is the output

anonimnystefy · 2013-09-21 05:30:09

The definition of ans2 is incorrect.

bobbym · 2013-09-21 05:33:29

What would you do from there?

anonimnystefy · 2013-09-21 05:42:15

What did you try to do to get ans2, ie. what did you think Select[ans1, Length[Union[#[[All, 2]]]] == 4 &]; would do?

Math Is Fun Forum

#1 2013-09-20 13:16:14

Combinatorics

#2 2013-09-20 20:06:17

Re: Combinatorics

#3 2013-09-20 21:32:34

Re: Combinatorics

#4 2013-09-20 22:47:47

Re: Combinatorics

#5 2013-09-20 22:54:09

Re: Combinatorics

#6 2013-09-21 00:13:06

Re: Combinatorics

#7 2013-09-21 00:19:05

Re: Combinatorics

#8 2013-09-21 00:24:07

Re: Combinatorics

#9 2013-09-21 00:32:34

Re: Combinatorics

#10 2013-09-21 03:54:21

Re: Combinatorics

#11 2013-09-21 04:06:37

Re: Combinatorics

#12 2013-09-21 04:09:04

Re: Combinatorics

#13 2013-09-21 04:35:59

Re: Combinatorics

#14 2013-09-21 04:40:45

Re: Combinatorics

#15 2013-09-21 04:42:25

Re: Combinatorics

#16 2013-09-21 04:47:08

Re: Combinatorics

#17 2013-09-21 04:53:49

Re: Combinatorics

#18 2013-09-21 04:58:57

Re: Combinatorics

#19 2013-09-21 05:12:39

Re: Combinatorics

#20 2013-09-21 05:17:50

Re: Combinatorics

#21 2013-09-21 05:20:54

Re: Combinatorics

#22 2013-09-21 05:24:55

Re: Combinatorics

#23 2013-09-21 05:30:09

Re: Combinatorics

#24 2013-09-21 05:33:29

Re: Combinatorics

#25 2013-09-21 05:42:15

Re: Combinatorics

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