Stumped: Can't Solve the Problem After Multiple Tries

[sahilmm15]

Homework Statement

The number of 6-digit numbers that can be made with the digits 0,1,2,3,4 and 5 so that even digits occupy odd places, is?

Relevant Equations

N/A

I cannot attempt the problem after multiple tries. Can you give me an insight to the problem.

 

[Delta2]

you have three even digits (I suppose we can take 0 as an even) {0,2,4 }and three odd digits {1,3,5}. And you want the total number of 6-digit numbers so that even digits occupy odd places AND odd digits occupy even places.

If we make the problem simpler and I ask for the total number of 2 (two) digits numbers with the same 6 possible digits, such that even digits are on the odd place (first place) and odd digits are on the even place (second place) how would you proceed?

 

[sahilmm15]

you have three even digits (I suppose we can take 0 as an even) {0,2,4 }and three odd digits {1,3,5}. And you want the total number of 6-digit numbers so that even digits occupy odd places AND odd digits occupy even places.

If we make the problem simpler and I ask for the total number of 2 (two) digits numbers with the same 6 possible digits, such that even digits are on the odd place (first place) and odd digits are on the even place (second place) how would you proceed?

6 I guess?

 

[Delta2]

No they are not 6 in total they are 9 but tell me how did you think to form the answer 6?

 

[sahilmm15]

No they are not 6 in total they are 9 but tell me how did you think to form the answer 6?

6 because we cannot put 0 in the first place else the number would be single digit??

 

[Delta2]

Ok I see , you might be right but it depends on the interpretation of the problem statement whether we are allowed to put 0 in the first place.

Anyway let's suppose that we can't put 0 in first place as you say, then the answer is indeed 6. How would you proceed to generalize it for 3 ,4 ,5 or 6 places

 

[sahilmm15]

Ok I see , you might be right but it depends on the interpretation of the problem statement whether we are allowed to put 0 in the first place.

Anyway let's suppose that we can't put 0 in first place as you say, then the answer is indeed 6. How would you proceed to generalize it for 3 ,4 ,5 or 6 places

This doubt is cleared. Now my only doubt remains is whether we can repeat the digits or not? By seeing the question what do you think? I think it should not be repetitive because it mentioned We need to form a six digit no by these digits. So I think we cannot repeat, what do you think??

 

[Delta2]

Well I am not sure to be honest, that was one question I had myself too, whether we can repeat or not.

 

[sahilmm15]

Well I am not sure to be honest, that was one question I had myself too, whether we can repeat or not.

Wait I think I got it. it says No of 6 digit numbers that can be made with the digits 0,1,2,3,4,5. Ok? But if we repeat the digits, like one number is 250143 ,after repeating say 250123 we are not going according to the question then . We need to make a six digit number with 0,1,2,3,4,5(everything included). So we are violating the question. Hence, we cannot repeat??

 

[Delta2]

Wait I think I got it. it says No of 6 digit numbers that can be made with the digits 0,1,2,3,4,5. Ok? But if we repeat the digits, like one number is 250143 ,after repeating say 250123 we are not going according to the question then . We need to make a six digit number with 0,1,2,3,4,5(everything included). So we are violating the question. Hence, we cannot repeat??

It all depends on the interpretation of the problem's statement. What you saying is one possible interpretation but I THINK it's not the only valid interpretation possible.

 

[sahilmm15]

It all depends on the interpretation of the problem's statement. What you saying is one possible interpretation but I THINK it's not the only valid interpretation possible.

Hmm but the correct answer only allows not repeating. But your point is valid.

 

[PeroK]

Homework Statement:: The number of 6-digit numbers that can be made with the digits 0,1,2,3,4 and 5 so that even digits occupy odd places, is?
Relevant Equations:: Fundamental Principle of counting.

I cannot attempt the problem after multiple tries. Can you give me an insight to the problem.

It seems clear that each digit appears once. Otherwise, what is and isn't allowed is ambiguous and it could get very complicated.

You say you can't attempt the problem, but yet it is a trivial matter to start writing out the solutions. And, as soon as you do that, you should see the pattern. Assuming the leftmost digit is the first place, then we start with:

012345

Then, we have to decide how to go through this methodically. Let's fix the odd numbers and start moving the even numbers around. The next ones might be:

014325
210345
214305

And it should soon be obvious how to count them all.

I think you need to ask yourself why you didn't think of this.

 

[sahilmm15]

It seems clear that each digit appears once. Otherwise, what is and isn't allowed is ambiguous and it could get very complicated.

You say you can't attempt the problem, but yet it is a trivial matter to start writing out the solutions. And, as soon as you do that, you should see the pattern. Assuming the leftmost digit is the first place, then we start with:

012345

Then, we have to decide how to go through this methodically. Let's fix the odd numbers and start moving the even numbers around. The next ones might be:

014325
210345
214305

And it should soon be obvious how to count them all.

I think you need to ask yourself why you didn't think of this.

Thanks for your help but I solved the problem after some 5 minutes but one doubt remained which we ( Me and Delta2 were discussing). whether we can repeat the digits or not? By seeing the question what do you think? I think it should not be repetitive because it mentioned We need to form a six digit no by these digits. So I think we cannot repeat, what do you think?? Also I think in your answer numbers starting from 0 won't count. Because then it would no longer remain 6 digit.

 

[PeroK]

Thanks for your help but I solved the problem after some 5 minutes but one doubt remained which we ( Me and Delta2 were discussing). whether we can repeat the digits or not? By seeing the question what do you think? I think it should not be repetitive because it mentioned We need to form a six digit no by these digits. So I think we cannot repeat, what do you think?? Also I think in your answer numbers starting from 0 won't count. Because then it would no longer remain 6 digit.

It's too complicated if you allow repetition. It's not even clear whether it means "all odd places must have even digits" or "any even digit must be in an odd place".

It guess it should have said, but by default keep it simple. I must admit I assumed they were talking about permutations of the digits.

 

[sahilmm15]

It's too complicated if you allow repetition. It's not even clear whether it means "all odd places must have even digits" or "any even digit must be in an odd place".

It guess it should have said, but by default keep it simple. I must admit I assumed they were talking about permutations of the digits.

Hmm I agree with you. It would be too complicated then regarding high school standards.

 

[Delta2]

Ehm, sorry I don't understand why it would be too complicated if we allow repetition. Wouldn't simply be 3x3x3x3x3x3=3^6?

 

[Delta2]

I guess the interpretation of the problem becomes complicated if we are allowed for repetitions because then there are too many possible interpretations.

 

[Delta2]

Then to complicate it he played this trick lol.

Huh, what trick?

 

[sahilmm15]

I just solved another problem from the same book. It completely tricked me. Worth a try The number of distinct rational numbers x such that 0<x<1 and x=p/q, where p,q belongs to {1,2,3,4,5,6} is ?

 

[Delta2]

Is it 5+4+3+2+1=5(5+1)/2=15
?

 

[sahilmm15]

Is it 5+4+3+2+1=5(5+1)/2=15
?

Sir, you got tricked in the same way I got. That's the beauty.

 

[Delta2]

e hehe, that's ok i now understood, they are fewer than 15 cause i double counted some. for example 1/2 and 2/4

 

[sahilmm15]

Sir, you got tricked in the same way I got. That's the beauty.

1/2, 1/3, 1/4, 1/5, 1/6 , 2/3 2/4=1/2= numbers repeated,... so on.

 

[sahilmm15]

e hehe, that's ok i now understood, they are fewer than 15 cause i double counted some. for example 1/2 and 2/4

These kind of problems just forces people to think in different terms.

 

[haruspex]

Ehm, sorry I don't understand why it would be too complicated if we allow repetition. Wouldn't simply be 3x3x3x3x3x3=3^6?

Wouldn't it be 2x6x3x6x3x6?

 

[Delta2]

Wouldn't it be 2x6x3x6x3x6?

Yes it can be that as well, depends what the problem wants from us, unfortunately the problem statement is a bit short in details, would need to tell us if it allows repetitions and if even places can have only odd digits or not.

 

[sahilmm15]

Yes it can be that as well, depends what the problem wants from us, unfortunately the problem statement is a bit short in details, would need to tell us if it allows repetitions and if even places can have only odd digits or not.

I think the problem allows all possible interpretations, but author wants us to think in simplest terms.