Counting exercise

[Nube]

Hi, I don't know how to solve this:
- How many ways can you fill 10 glasses ordered and numbered from 1 to 10 using
4 possible drinks so that no two consecutive glasses are left with the same drink?
Suppose each glass is filled with some drink.

At first glance I would guess that it is [imath]4^{10}[/imath] by product rule, but I don't know how to make it so that no two consecutive glasses are left with the same drink. Thank you very much

 

[Cubist]

If I fill glass 1 with orange juice, then how many ways can you fill glass 2?

 

[Nube]

If I fill glass 1 with orange juice, then how many ways can you fill glass 2?

I guess 3, right?

 

[pka]

Hi, I don't know how to solve this:
- How many ways can you fill 10 glasses ordered and numbered from 1 to 10 using
4 possible drinks so that no two consecutive glasses are left with the same drink?
Suppose each glass is filled with some drink.

At first glance I would guess that it is [imath]4^{10}[/imath] by product rule, but I don't know how to make it so that no two consecutive glasses are left with the same drink. Thank you very much

Question: Do you have to use at least one of each kind?
If not we could have [imath]OLOLOLOLOL[/imath] All either orange squash or lime sofa.

 

[Nube]

Pregunta : ¿Tienes que usar al menos uno de cada tipo?
Si no, podríamos tener [imath]OLOLOLOLOL[/imath] Todos, ya sea calabaza naranja o sofá lima.

The condition is that you can't have an empty glass, it doesn't say anything about using at least one of each type. I add that the result is 78732, but I don't know how to get there.

 

[Cubist]

I guess 3, right?

Yes.

Then, for each of those three ways, how many ways can you fill glass #3?
And then think about the same question for glass #4 etc... does this give you any ideas?

 

[Nube]

Yes.

Then, for each of those three ways, how many ways can you fill glass #3?
And then think about the same question for glass #4 etc... does this give you any ideas?

So far I have come , intuitively I would say for vessel #3 it is 2, but I don't see it. And I would not find the result following this logic either.

 

[pka]

The condition is that you can't have an empty glass, it doesn't say anything about using at least one of each type. I add that the result is 78732, but I don't know how to get there.

Well [imath]4\cdot 3^9=78732[/imath] Can you explain?

 

[Nube]

Well [imath]4\cdot 3^9=78732[/imath] Can you explain?

No, that's why I made the post, I have the solution because in the list of exercises I had the solution, but not the procedure.

 

[JeffM]

Please go back to Cubist’s questions.

How many ways can you fill glass 1?

How many ways can you fill glass 2?

In general, if k > 1, how many ways can you fill glass k+1?

 

[pka]

If you think that we are going all the work for you then you are mistaken.
You must try and explain the equation I posted.
Where does the 4 come from & why?
What the 3? And the 9??

 

[Cubist]

Please go back to Cubist’s questions.

How many ways can you fill glass 1?

How many ways can you fill glass 2?

In general, if k > 1, how many ways can you fill glass k+1?

...assume that glass k+2 is empty. In other words, the glasses are being filled in ascending number order

EDIT: that's addressed to @Nube , I'm just adding extra explanation to the post from @JeffM

 

[Nube]

If you think that we are going all the work for you then you are mistaken.
You must try and explain the equation I posted.
Where does the 4 come from & why?
What the 3? And the 9??

What equation? I only see a multiplication, where is the equality and the unknown?
Anyway, I guess the 4 is for the number of ways the first glass can be filled and the 3^{9} is the ways the remaining glasses can be filled.

 

[JeffM]

What equation? I only see a multiplication, where is the equality and the unknown?
Anyway, I guess the 4 is for the number of ways the first glass can be filled and the 3^{9} is the ways the remaining glasses can be filled.

That is correct.

Can you explain in words why, for ABSOLUTELY EVERY GLASS AFTER THE FIRST, there are exactly three ways to fill that glass. If you cannot articulate that, you do not understand the answer.

By the way, if by “procedure“ you mean formula, combinatorics is not primarily a subject that requires memorizing formulas; it is a subject that applies a few simple formulas after very careful thuoght.

 

[Nube]

That is correct.

Can you explain in words why, for ABSOLUTELY EVERY GLASS AFTER THE FIRST, there are exactly three ways to fill that glass. If you cannot articulate that, you do not understand the answer.

By the way, if by “procedure“ you mean formula, combinatorics is not primarily a subject that requires memorizing formulas; it is a subject that applies a few simple formulas after very careful thuoght.

That's what I don't understand, I understand that in order for the same drink not to appear next to the first glass "one drink would be removed".
And by procedure I mean how the result or reasoning is arrived at. I apologize if you do not understand me, because of my bad English, I speak Spanish :c

 

[pka]

That's what I don't understand, I understand that in order for the same drink not to appear next to the first glass "one drink would be removed".
And by procedure I mean how the result or reasoning is arrived at. I apologize if you do not understand me, because of my bad English, I speak Spanish :c

Maybe a visual will help you. There is one of four that can go into place 1: [imath]L,O,C,\text{ or }A[/imath]
[imath]\boxed{\underbrace {\quad \quad }_1}\boxed{\underbrace {\quad \quad }_2}\boxed{\underbrace {\quad \quad }_3}\boxed{\underbrace {\quad \quad }_4}\boxed{\underbrace {\quad \quad }_5}\boxed{\underbrace {\quad \quad }_6}\boxed{\underbrace {\quad \quad }_7}\boxed{\underbrace {\quad \quad }_8}\boxed{\underbrace {\quad \quad }_9}\boxed{\underbrace {\quad \quad }_{10}}[/imath]
Suppose we choose [imath]\bf L[/imath]ime to go into glass #1. Then there are three left that can be used in 'glass #2.
Say we next choose [imath]\bf C[/imath]ola to be used in 'glass #2. Now again there are three left that can be used in 'glass #3.
Thus by repeating the process of choosing eight times we get [imath]4\cdot 3^9[/imath]

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

 

[Nube]

Tal vez una imagen te ayude. Hay uno de los cuatro que pueden ir en el lugar 1: [imath]L,O,C,\text{ o }A[/imath]
[imath]\boxed{\underbrace {\quad \quad }_1}\boxed{\underbrace {\quad \quad }_2}\boxed{\underbrace {\quad \quad }_3}\boxed{\underbrace {\quad \quad }_4}\boxed{\underbrace {\quad \quad }_5}\boxed{\underbrace {\quad \quad }_6}\boxed{\underbrace {\quad \quad }_7}\boxed{\underbrace { \quad \quad }_8}\boxed{\underbrace {\quad \quad }_9}\boxed{\underbrace {\quad \quad }_{10}}[/imath]
Supongamos que elegimos [imath]\bf L[/imath]ime para entrar en el vaso #1. Luego quedan tres que se pueden usar en el 'vaso #2.
Digamos que a continuación elegimos [imath]\bf C[/imath]ola para usar en el 'vaso #2. Ahora nuevamente quedan tres que se pueden usar en el 'vaso #3.
Por lo tanto, al repetir el proceso de elegir ocho veces obtenemos [imath]4\cdot 3^9[/imath]

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

I think I understand
For the first glass: We have 4 options, as we have no restrictions, we assume we choose "L".
For the second glass: We have 4 options, but there can't be another "L" so we are left with 3 options, we assume we choose "C".
For the third cup: We have 4 options, but in this case we cannot choose "C" so we are left with 3 options.
Is this correct?

 

[pka]

I think I understand
For the first glass: We have 4 options, as we have no restrictions, we assume we choose "L".
For the second glass: We have 4 options, but there can't be another "L" so we are left with 3 options, we assume we choose "C".
For the third cup: We have 4 options, but in this case we cannot choose "C" so we are left with 3 options.
Is this correct?

Correct.

 

[Steven G]

Yes, you are correct. So what do you think the final answer is?

 

[Nube]

Yes, you are correct. So what do you think the final answer is?

According to the product rule, there are 78732 ways to fill 10 glasses numbered 1 to 10 using 4 different drinks without two glasses having consecutively two drinks.