Permutations Questions: how many 6-digit binaries can be formed? how many are palindromes?

[yobacul]

Can someone please tell me if I reasoned it out correctly? Thanks

1) How many different 6-digit binary numbers (using only digits 0 and 1) can be formed?
2) How many 6-digit numbers are palindrome?
3) What is the probability that a 6 digit number chosen at random is a palindrome starting with digit 1?

1) Am I right to say the answer is 64 (2 to the power of 6)? My only problem is can a binary number consist of zeros only or ones only?
2) For this my answer is 8, as there are 2x2x2x1x1x1=8 ways of getting a palindrome
3) half of the palindromes in 2) start with a 1, so this answer is 4/64=1/16.

 

[yobacul]

A box contains 4 white disks and 12 black disks. The disks are taken out one by one at random from the box and placed on top of each other. How many different arrangements are possible such that the bottom 3 disks have alternating colours?

#My reasoning and answer

(4x12x3x13!)/(11!x2!)+(12x4x11x13!)/(10!x3!)= 464,256

 

[pka]

Can someone please tell me if I reasoned it out correctly? Thanks

1) How many different 6-digit binary numbers (using only digits 0 and 1) can be formed?
2) How many 6-digit numbers are palindrome?
3) What is the probability that a 6 digit number chosen at random is a palindrome starting with digit 1?

1) Am I right to say the answer is 64 (2 to the power of 6)? My only problem is can a binary number consist of zeros only or ones only?
2) For this my answer is 8, as there are 2x2x2x1x1x1=8 ways of getting a palindrome
3) half of the palindromes in 2) start with a 1, so this answer is 4/64=1/16.

Your answers are correct. Note that any three digit binary number (there are eight) can be reversed to create a palindrome.
You correctly say that half of those begin with a 1.

 

[Steven G]

I'm sorry but I don't agree with some of your answers/reasons.
Yes, it is true that problem 1 talks about using only binary numbers.
Problems 2 and 3 does not mention that the six digits are each binary.
I therefore do not agree with the answer for 2 and 3.

 

[Dr.Peterson]

I'm sorry but I don't agree with some of your answers/reasons.
Yes, it is true that problem 1 talks about using only binary numbers.
Problems 2 and 3 does not mention that the six digits are each binary.
I therefore do not agree with the answer for 2 and 3.

The problem could be stated more clearly, but I think it is reasonable to interpret it as saying that all three parts are asking about the same kind of number: 6-digit binary.

It would be not just poor wording, but meanness to switch to decimal numbers without saying so -- or, worse, to intentionally make the last two parts unsolvable because the base is unknown.

Normal communication involves recognizing that all facts don't have to be restated in every sentence, and that if a fact (the base) is needed in order to understand something, then it is reasonable to assume it carries over.

@nad081, I agree that your reasoning is all good. As for your second question ... that should have been in a separate thread.

 

[pka]

A box contains 4 white disks and 12 black disks. The disks are taken out one by one at random from the box and placed on top of each other. How many different arrangements are possible such that the bottom 3 disks have alternating colours?

There are two possible cases for the bottom three.
[imath]W~B~W\text{ or }B~W~B[/imath]
In the first case that leaves [imath]2\text{ white disks and }11\text{ black disks.}[/imath]
In the second case that leaves [imath]3\text{ white disks and }10\text{ black disks.}[/imath]

JUST FOR EXAMPLE> Suppose that we have six Red disks and four Green disks.
Those ten disks can be arranged in [imath]\dfrac{10!}{4!\cdot 6!}[/imath] ways.
Can you finish?
[imath][/imath][imath][/imath]

 

[Dr.Peterson]

A box contains 4 white disks and 12 black disks. The disks are taken out one by one at random from the box and placed on top of each other. How many different arrangements are possible such that the bottom 3 disks have alternating colours?

#My reasoning and answer

(4x12x3x13!)/(11!x2!)+(12x4x11x13!)/(10!x3!)= 464,256

You haven't explained your reasoning, but it looks like you may be considering the disks of a given color to be distinguishable, since your numerical answer is so large. You should consider them indistinguishable. Only the colors count, not which disks they are.

If this doesn't help, please tell us in words why you multiplied the numbers you did.

 

[Steven G]

The problem would have been clear if it said:
The digits are all binary digits.
1) How many different 6-digit numbers can be formed?
2) How many 6-digit numbers are palindrome?
3) What is the probability that a 6 digit number chosen at random is a palindrome starting with digit 1?

 

[Steven G]

...or, worse, to intentionally make the last two parts unsolvable because the base is unknown.

How many 6-digit numbers are palindrome?
It seems clear to me what 6-digit numbers are. The base is 10. When ever someone asks you to write down a 6 digit number it is always assumed base 10 unless otherwise stated. To say that this problem can't be done because the base is unknown is strange to say the least. After all, either it is base 2 or base 10.

From what I learned is when you are given three problems, (not 1a, 1b and 1c, but rather 1, 2 and 3), they are all independent of one another.


Problem number two is quite doable. The question is rather a 6-digit number can start with a 0 or not.

Note that abccba is a 6 digit palindrome. That is, the 1st 3 digits determine the 6 digit palindrome. The number of 6 digit palindrome will be 9*10^2, if 0 can't be the 1st digit, and 10^3 if 0 can be the 1st digit.

The OP said that the answer to 2 is 2x2x2x1x1x1. Just because a number is binary does not suddenly allow the 1st digit to be 0. That needs to be specified.

 

[Dr.Peterson]

It seems clear to me what 6-digit numbers are. The base is 10.

Nonsense. Context matters.

From what I learned is when you are given three problems, (not 1a, 1b and 1c, but rather 1, 2 and 3), they are all independent of one another.

Maybe. Not always. But we haven't see the original form of the problems, so I wouldn't be so sure.

Problem number two is quite doable. The question is rather a 6-digit number can start with a 0 or not.

Do you mean "whether"? Please proofread.

But this is a valid question. As already noted, the problem (if copied accurately, which I doubt) is not well written.

And this issue affects even question 1! I might argue that binary numbers commonly are in a computer context, where it's storage space, not content, that determines the stated digit count, but that's hard to say without knowing the actual context. Generally, however, if a student implies a certain interpretation, I go with that, figuring they know from their context how terms are used, unless there's good reason to disagree with them. And here it's clear that the student took the problem(s) to be all about binary numbers, and to allow the first digit to be 0. (And question 3 strongly suggests the latter, as it considers the first digit being 1 to be a restriction.)

If the student now wrote back and said his answers don't agree with the book's, we'd definitely be pointing out these ambiguities.

@nad081, please show us an image of the actual problem, and telling us the context (in particular, whether you have been told how to interpret "6-digit [binary] number").

 

[yobacul]

Your answers are correct. Note that any three digit binary number (there are eight) can be reversed to create a palindrome.
You correctly say that half of those begin with a 1.

Thanks a lot for taking the time to check them. Appreciated

 

[yobacul]

You haven't explained your reasoning, but it looks like you may be considering the disks of a given color to be distinguishable, since your numerical answer is so large. You should consider them indistinguishable. Only the colors count, not which disks they are.

If this doesn't help, please tell us in words why you multiplied the numbers you did.

You haven't explained your reasoning, but it looks like you may be considering the disks of a given color to be distinguishable, since your numerical answer is so large. You should consider them indistinguishable. Only the colors count, not which disks they are.

If this doesn't help, please tell us in words why you multiplied the numbers you did.

 

[yobacul]

Thanks! You’re right there. I considered the disks to be distinguishable. Will give it another go.

 

[yobacul]

Nonsense. Context matters.


Maybe. Not always. But we haven't see the original form of the problems, so I wouldn't be so sure.


Do you mean "whether"? Please proofread.

But this is a valid question. As already noted, the problem (if copied accurately, which I doubt) is not well written.

And this issue affects even question 1! I might argue that binary numbers commonly are in a computer context, where it's storage space, not content, that determines the stated digit count, but that's hard to say without knowing the actual context. Generally, however, if a student implies a certain interpretation, I go with that, figuring they know from their context how terms are used, unless there's good reason to disagree with them. And here it's clear that the student took the problem(s) to be all about binary numbers, and to allow the first digit to be 0. (And question 3 strongly suggests the latter, as it considers the first digit being 1 to be a restriction.)

If the student now wrote back and said his answers don't agree with the book's, we'd definitely be pointing out these ambiguities.

@nad081, please show us an image of the actual problem, and telling us the context (in particular, whether you have been told how to interpret "6-digit [binary] number").

 

[yobacul]

So the second question should have read ‘how many of these 6 digit numbers are binary? So second and third questions build on the first

 

[yobacul]

There are two possible cases for the bottom three.
[imath]W~B~W\text{ or }B~W~B[/imath]
In the first case that leaves [imath]2\text{ white disks and }11\text{ black disks.}[/imath]
In the second case that leaves [imath]3\text{ white disks and }10\text{ black disks.}[/imath]

JUST FOR EXAMPLE> Suppose that we have six Red disks and four Green disks.
Those ten disks can be arranged in [imath]\dfrac{10!}{4!\cdot 6!}[/imath] ways.
Can you finish?
[imath][/imath][imath][/imath]

yes yes, thank you. I was making the mistake of considering the disks as if they were distinguishable. So the answer is [imath]\dfrac{13!}{11!\cdot 2!}+\dfrac{13!}{10!\cdot 3!} ways.[/imath]

 

[yobacul]

So the second question should have read ‘how many of these 6 digit binary numbers are palindrome? So second and third questions build on the first

 

[yobacul]

There are two possible cases for the bottom three.
[imath]W~B~W\text{ or }B~W~B[/imath]
In the first case that leaves [imath]2\text{ white disks and }11\text{ black disks.}[/imath]
In the second case that leaves [imath]3\text{ white disks and }10\text{ black disks.}[/imath]

JUST FOR EXAMPLE> Suppose that we have six Red disks and four Green disks.
Those ten disks can be arranged in [imath]\dfrac{10!}{4!\cdot 6!}[/imath] ways.
Can you finish?
[imath][/imath][imath][/imath]

 

[Dr.Peterson]

So the second question should have read ‘how many of these 6 digit numbers are binary? So second and third questions build on the first

No, that's nonsense. Surely you don't mean that!

But, yes, they are (in my view) related.

yes yes, thank you. I was making the mistake of considering the disks as if they were distinguishable. So the answer is [imath]\dfrac{13!}{11!\cdot 2!}+\dfrac{13!}{10!\cdot 3!} ways.[/imath]

Yes. You have to choose either 2 or 3 of the remaining 13 locations to hold a white disk.

(By the way, do you see yet why the second problem should have been in a separate thread? I'm getting dizzy trying to decide which posts are about which problem.)

 

[stapel]

Note: User "nad081" changed username to "yobacul".

Change approved.

Eliz.