As part of his act, a magician has one red chip, one green chip, and one blue chip. He does a trick in which all three of those chips...

[eddy2017]

As part of his act, a magician has one red chip, one green chip, and one blue chip. He does a trick in which all three of those chips are placed in a row on a table. considering all possible arrangements of those chips, in how many of these arrangements will the red chip and green chip be side by side.

I have watched several videos on permutation and combination at my level.
I know that by going by the list option I can have the answer:
RBG - RGB - BRG - BGR - GAR - GRB
so, taking a look at the options I realize that there are 4 arrangements of the type they ask for.
but then, I wonder that if I want to go the formula way, then I should go with
nPr = n!/ (n-r)!
but using this the result is 6.
my question is, is there any formula that might give me a direct solution, like, 4 in this case.
thank you,
eddy
Note: I edited the letters from Spanish to English.

 

[lev888]

As part of his act, a magician has one red chip, one green chip, and one blue chip. He does a trick in which all three of those chips are placed in a row on a table. considering all possible arrangements of those chips, in how many of these arrangements will the red chip and green chip be side by side.

I have watched several videos on permutation and combination at my level.
I know that by going by the list option I can have the answer:
RBG - RGB - BRG - BGR - GAR - GRB
so, taking a look at the options I realize that there are 4 arrangements of the type they ask for.
but then, I wonder that if I want to go the formula way, then I should go with
nPr = n!/ (n-r)!
but using this the result is 6.
my question is, is there any formula that might give me a direct solution, like, 4 in this case.
thank you,
eddy
Note: I edited the letters from Spanish to English.

I would use this approach.
1. Consider RG as one chip, use the formula to count the number of all arrangements RG and B.
2. For each of the arrangement above we can rearrange RG in how many way?
3. Therefore, the total is ?

 

[eddy2017]

In 4 ways only if I consider RG a chip and then B. Considering RG has to be side by side.
That is clear but I wonder why the formula does not work here?. That is what puzzles me. It gives me 6 arrangements and then I have to go and select the ones that apply. Don't know, lev, if you know what I mean. It does not give 4 as a result. I have to go and discern it is 4.

 

[lev888]

In 4 ways only if I consider RG a chip and then B. Considering RG has to be side by side.
That is clear but I wonder why the formula does not work here?. That is what puzzles me. It gives me 6 arrangements and then I have to go and select the ones that apply. Don't know, lev, if you know what I mean. It does not give 4 as a result. I have to go and discern it is 4.

A formula is a recipe. It takes specific ingredients and produces a specific dish. If you want a turkey you don't use an apple pie recipe.
What's the description of the formula you used?

 

[pka]

In 4 ways only if I consider RG a chip and then B. Considering RG has to be side by side.
That is clear but I wonder why the formula does not work here?. That is what puzzles me. It gives me 6 arrangements and then I have to go and select the ones that apply. Don't know, lev, if you know what I mean. It does not give 4 as a result. I have to go and discern it is 4.

While I agree that formula is not the correct term, I will give you an larger example.
There are seven colours in a rainbow: Red, Orange, Yellow, Green, Blue, Indigo, & Violet..
Now there are [imath]7!=5040[/imath] ways to arrange the letters [imath]\{\bf{R,O,Y,G,B,I,V}\}[/imath] that is simplify using a factorial.
Suppose I need count the number of ways in which the [imath]\bf {R~\&~V}[/imath] are next to each other.
Well I imagine gluing them together as in [imath]\boxed{RV},O,Y,G,B,I[/imath] Now there are six to arrange.
[imath]6!=720[/imath] but the real answer is [imath]2\cdot 6!=1440[/imath]. Where did the two come from?

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[eddy2017]

A formula is a recipe. It takes specific ingredients and produces a specific dish. If you want a turkey you don't use an apple pie recipe.
What's the description of the formula you used?

Here's the description as I have seen it explained in almost every source I have checked
Permutation: nPr represents the probability of selecting an ordered set of ‘r’ objects from a group of ‘n’ number of objects. The order of objects matters in case of permutation. The formula to find nPr is given by:

nPr = n!/(n-r)!
So in our case n would be 3 the number of chips, and then r would be ...I would say 2 but I am not so sure now what r should be since I am looking for green and blue to be next to each other.

 

[eddy2017]

While I agree that formula is not the correct term, I will give you an larger example.
There are seven colours in a rainbow: Red, Orange, Yellow, Green, Blue, Indigo, & Violet..
Now there are [imath]7!=5040[/imath] ways to arrange the letters [imath]\{\bf{R,O,Y,G,B,I,V}\}[/imath] that is simplify using a factorial.
Suppose I need count the number of ways in which the [imath]\bf {R~\&~V}[/imath] are next to each other.
Well I imagine gluing them together as in [imath]\boxed{RV},O,Y,G,B,I[/imath] Now there are six to arrange.
[imath]6!=720[/imath] but the real answer is [imath]2\cdot 6!=1440[/imath]. Where did the two come from?

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

Shouldn't RV then be counted as 1
so, n= 6 and r=1?

I think the Permutation formula or whatever we may call it doesn't work here as it is.

 

[eddy2017]

let me correct what I have previously said. I think the permutation formula is correct to use but the choosing has to be done by hand so to speak, I see no other way.

 

[pka]

Shouldn't RV then be counted as 1
so, n= 6 and r=1?

i think the Permutation formula or whatever we may call it doesn't work here as it is.

One last time: THERE IS NO FORMULA! I you cannot give that idea up then I must quit answering with you.
The is only your brain. If were have ten letters [imath]\{A,B.C,D,E,F,G,H,J,K\}[/imath] they can form a que in [imath]10!=3628800[/imath] ways.
There is no formula to that just knowing that we can place the A is ten places, the B in nine places. the C in eight places, etc.
In how many of those are the A, B, C all together? The answer is [imath](3!)(8!)[/imath]
Now you tell us why that is the answer. There is no formula to it.

 

[eddy2017]

One last time: THERE IS NO FORMULA! I you cannot give that idea up then I must quit answering with you.
The is only your brain. If were have ten letters [imath]\{A,B.C,D,E,F,G,H,J,K\}[/imath] they can form a que in [imath]10!=3628800[/imath] ways.
There is no formula to that just knowing that we can place the A is ten places, the B in nine places. the C in eight places, etc.
In how many of those are the A, B, C all together? The answer is [imath](3!)(8!)[/imath]
Now you tell us why that is the answer. There is no formula to it.

Understood, pka. Thanks a lot!.

 

[lev888]

Here's the description as I have seen it explained in almost every source I have checked
Permutation: nPr represents the probability of selecting an ordered set of ‘r’ objects from a group of ‘n’ number of objects. The order of objects matters in case of permutation. The formula to find nPr is given by:

nPr = n!/(n-r)!
So in our case n would be 3 the number of chips, and then r would be ...I would say 2 but I am not so sure now what r should be since I am looking for green and blue to be next to each other.

There is a lot of confusion going on here.
First, you mentioned probability. Please clarify. Where does probability come into the picture?

Second, your formula is for counting the number of ways to select r objects out of n objects. This is not at all what you need to count in this problem. You have 3 objects and you are counting ways to order them.
Let's simplify the problem by removing the part about R and G being next to each other. Can you solve it without listing all combinations? What's the formula?

 

[eddy2017]

Let's simplify the problem by removing the part about R and G being next to each other. Can you solve it without listing all combinations? What's the formula?

3! then,
3*2*1 =6

 

[JeffM]

It may be an error to intrude where skilled mathematicians have already answered. But then my wife has long claimed that I am an idiot.

I agree with pka that formulas are no substitute for thought.

I agree with lev that a simple way to think about this problem is to think about a red and green chip glued together side by side. So now we have two chips, one that is not blue and one that is blue.How many different ways can you put them side by side?

How many different ways can you put red and green side by side?

So what is the answer derived from counting principles?

What is the answer derived from listing?

 

[blamocur]

Let's simplify the problem by removing the part about R and G being next to each other. Can you solve it without listing all combinations? What's the formula?

3! then,
3*2*1 =6

That is right. Now subtract the ones where R and G are not next to each other, and you should get the same answer.

 

[eddy2017]

4

 

[lev888]

4

I'm guessing you counted 2 cases where R and G are not next to each other. What if in addition to R and G we had 50 other chips? Then counting all those cases would not be feasible.
My intent with the simplified problem was not to show you the way to a solution. It was to answer the question about the formula:
"my question is, is there any formula that might give me a direct solution, like, 4 in this case."

For the simple problem there is a formula: n!
Here the situation matches exactly what the formula assumes: we are arranging n distinct items. The number of such arrangements is n!

But if you introduce additional details, such as 2 items need always be next to each other, you are changing the dish - the same recipe can no longer be used. Is there a formula for the new situation? Who knows. There might be, if the new problem occurs often enough making it worthwhile for people to sit down and derive a formula for it. In this case I would say it's unlikely. So, you need to think about the problem and come up with either your own "direct" formula or a multi-step solution that makes use of several known formulas. In post number 2 I outlined one such solution.