P&c q2

[Saumyojit]

Given 5 different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye?


I approached the sum like this :

I already picked one green and one blue .

There are nine total green blue dyes.

Choosing two out of 9 --> 5c1*4c1

After choosing two dyes one from green and one blue.

So now I have 10 dyes.

Each have a option of yes or no .
So, 2^10 =1024

1024* 5c1*4c1= ....

As far as I broke down some combination ,
G1B1 can happen

G1B1 G2 B2

G1 B1 r1 all blue dyes ...




Don't give alternate solution

Where am I wrong.

 

[JeffM]

How do we even know that you are wrong? You have not told us what your answer is. Nor have you told us your reasoning in any detail. So how could we know where you went wrong if you did?

By the way. How many different dyes are you combining?

Two, three, four, what?

Hopefully, if we provide an alternative, it will be the correct answer. Why are you uninterested in that.

 

[Saumyojit]

Okay I worked on my approach and figured out that there are Combination being repeated .
That's why Answer is so Large

 

[Saumyojit]

How do we even know that you are wrong? You have not told us what your answer is.

By the way. How many different dyes are you combining?

Two, three, four, what?

With my approach many are being counted twice or thrice....

Can you modify my approach so that I get the correct answer.

Where it should be modified that's the main thing.

 

[JeffM]

With my approach many are being counted twice or thrice....

Can you modify my approach so that I get the correct answer.

Where it should be modified that's the main thing.

I am not even sure what the problem is. How many dyes are you combining?

 

[JeffM]

Are repetitions allowed?

 

[Dr.Peterson]

If you got this problem somewhere, please show the original statement (and where it came from), along with the answer if you have it.

If you made it up, then you have to state it fully, answering the questions that have been asked.

And you must show the specifics of your work if you want to ask questions about it.

Given 5 different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye?

Here is my attempt at clarifying this, which you can change as needed:

You have 5 G's, 4 B's, and 3 R's, all distinguishable. You have only one of each, so only one of each can be used. You want to mix together any number of them -- so you are choosing any subset of them -- subject to the condition that there must be at least one G and at least one B. In how many ways can this be done?

To my mind, the obvious method is subtraction. That's all I'll say.

I approached the sum like this :

I already picked one green and one blue .

There are nine total green blue dyes.

Choosing two out of 9 --> 5c1*4c1

After choosing two dyes one from green and one blue.

So now I have 10 dyes.

Each have a option of yes or no .
So, 2^10 =1024

1024* 5c1*4c1= ....

You are overcounting, as you said, because you can get the same mix by starting with a different pair but picking the original pair as part of the later choices. So if you continue this method, you will need to be able to remove duplicate combinations.

That may be possible, but it would take a lot more work than doing it by subtraction.

Can you modify my approach so that I get the correct answer.

Where it should be modified that's the main thing.

I'd modify it right at the start! Does that count?

How many possibilities are there that do not include at least one G and one B?

 

[pka]

Given five different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye?

Okay I worked on my approach and figured out that there are Combination being repeated .

You did not tell us how many total are many are being selected.
Moreover, those two posts are contradictory, because if the dyes of each colour are different as you posted, there cannot be any reparations.

 

[Saumyojit]

Okay I found another method to solve.
Pick any no of green=2^5 -1
Pick an no of blue =2^4 -1
Pick any red or Zero also= 2^3

Multiply each other to get 3720

 

[Dr.Peterson]

Very good (though you broke your rule and gave an alternate solution rather than modifying what you had).

And I get the same result by subtraction (actually inclusion-exclusion), with just a little more work than your multiplication method.