Permutations with alphabet
$begingroup$
A password must be created using only the lowercase letters of the alphabet. How many passwords can be created that are up to 9 lowercase letters in size?
For my answer, I have:
$26 * 27^8$
I am unsure if my answer is correct.
I have 27 as surely a NULL letter should be included? What I mean by NULL is that in case no letter is chosen (thus shortening the password size as it is up to 9 letters in size).
combinatorics discrete-mathematics permutations
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
$endgroup$
add a comment |
$begingroup$
A password must be created using only the lowercase letters of the alphabet. How many passwords can be created that are up to 9 lowercase letters in size?
For my answer, I have:
$26 * 27^8$
I am unsure if my answer is correct.
I have 27 as surely a NULL letter should be included? What I mean by NULL is that in case no letter is chosen (thus shortening the password size as it is up to 9 letters in size).
combinatorics discrete-mathematics permutations
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
$endgroup$
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us how you arrived at that answer. Then we may be able to help you. Perhaps start by answering the question with $2$ or $3$ instead of $9$ and an alphabet with fewer letters to see the pattern. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:50
$begingroup$
I don't think there is a NULL letter. Lowercase letters are usually defined to be ${a, b, c, dots, z}$. However, I think you do need to include the empty password (the password with $0$ letters).
$endgroup$
– Alex Vong
Dec 13 '18 at 17:59
$begingroup$
@AlexVong There are plenty of reasons to consider a null character (especially in programming) and there are even some times where it might help with calculations. This happens to not be one of those cases, however that does not mean that it does not have its uses elsewhere.
$endgroup$
– JMoravitz
Dec 13 '18 at 18:03
$begingroup$
@JMoravitz You're right. I should be more clear. What I meant to say is that the lowercase letters usually don't include the NULL character. Of course, we can have''in C!
$endgroup$
– Alex Vong
Dec 13 '18 at 18:07
$begingroup$
It seems like your answer would allow blanks in the middle such as "ab bd ", is that what you want? If the blanks must be at the end, then it might be easier to add up the numbers of words of lengths, 1, 2, 3, ... , 9 and leave the blanks out of the computation. Also, I wouldn't imagine the empty string is a legitimate password in this context but that's just based on what I think "password" usually means in our world, not some technical math definition.
$endgroup$
– Ned
Dec 13 '18 at 18:07
add a comment |
$begingroup$
A password must be created using only the lowercase letters of the alphabet. How many passwords can be created that are up to 9 lowercase letters in size?
For my answer, I have:
$26 * 27^8$
I am unsure if my answer is correct.
I have 27 as surely a NULL letter should be included? What I mean by NULL is that in case no letter is chosen (thus shortening the password size as it is up to 9 letters in size).
combinatorics discrete-mathematics permutations
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
$endgroup$
A password must be created using only the lowercase letters of the alphabet. How many passwords can be created that are up to 9 lowercase letters in size?
For my answer, I have:
$26 * 27^8$
I am unsure if my answer is correct.
I have 27 as surely a NULL letter should be included? What I mean by NULL is that in case no letter is chosen (thus shortening the password size as it is up to 9 letters in size).
combinatorics discrete-mathematics permutations
combinatorics discrete-mathematics permutations
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
edited Dec 14 '18 at 11:56
Key Flex
8,01761233
8,01761233
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
asked Dec 13 '18 at 17:45
Jasmine078Jasmine078
123
123
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us how you arrived at that answer. Then we may be able to help you. Perhaps start by answering the question with $2$ or $3$ instead of $9$ and an alphabet with fewer letters to see the pattern. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:50
$begingroup$
I don't think there is a NULL letter. Lowercase letters are usually defined to be ${a, b, c, dots, z}$. However, I think you do need to include the empty password (the password with $0$ letters).
$endgroup$
– Alex Vong
Dec 13 '18 at 17:59
$begingroup$
@AlexVong There are plenty of reasons to consider a null character (especially in programming) and there are even some times where it might help with calculations. This happens to not be one of those cases, however that does not mean that it does not have its uses elsewhere.
$endgroup$
– JMoravitz
Dec 13 '18 at 18:03
$begingroup$
@JMoravitz You're right. I should be more clear. What I meant to say is that the lowercase letters usually don't include the NULL character. Of course, we can have''in C!
$endgroup$
– Alex Vong
Dec 13 '18 at 18:07
$begingroup$
It seems like your answer would allow blanks in the middle such as "ab bd ", is that what you want? If the blanks must be at the end, then it might be easier to add up the numbers of words of lengths, 1, 2, 3, ... , 9 and leave the blanks out of the computation. Also, I wouldn't imagine the empty string is a legitimate password in this context but that's just based on what I think "password" usually means in our world, not some technical math definition.
$endgroup$
– Ned
Dec 13 '18 at 18:07
add a comment |
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us how you arrived at that answer. Then we may be able to help you. Perhaps start by answering the question with $2$ or $3$ instead of $9$ and an alphabet with fewer letters to see the pattern. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:50
$begingroup$
I don't think there is a NULL letter. Lowercase letters are usually defined to be ${a, b, c, dots, z}$. However, I think you do need to include the empty password (the password with $0$ letters).
$endgroup$
– Alex Vong
Dec 13 '18 at 17:59
$begingroup$
@AlexVong There are plenty of reasons to consider a null character (especially in programming) and there are even some times where it might help with calculations. This happens to not be one of those cases, however that does not mean that it does not have its uses elsewhere.
$endgroup$
– JMoravitz
Dec 13 '18 at 18:03
$begingroup$
@JMoravitz You're right. I should be more clear. What I meant to say is that the lowercase letters usually don't include the NULL character. Of course, we can have''in C!
$endgroup$
– Alex Vong
Dec 13 '18 at 18:07
$begingroup$
It seems like your answer would allow blanks in the middle such as "ab bd ", is that what you want? If the blanks must be at the end, then it might be easier to add up the numbers of words of lengths, 1, 2, 3, ... , 9 and leave the blanks out of the computation. Also, I wouldn't imagine the empty string is a legitimate password in this context but that's just based on what I think "password" usually means in our world, not some technical math definition.
$endgroup$
– Ned
Dec 13 '18 at 18:07
1
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us how you arrived at that answer. Then we may be able to help you. Perhaps start by answering the question with $2$ or $3$ instead of $9$ and an alphabet with fewer letters to see the pattern. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:50
$begingroup$
Welcome to stackexchange. Please edit the question to show us how you arrived at that answer. Then we may be able to help you. Perhaps start by answering the question with $2$ or $3$ instead of $9$ and an alphabet with fewer letters to see the pattern. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:50
$begingroup$
I don't think there is a NULL letter. Lowercase letters are usually defined to be ${a, b, c, dots, z}$. However, I think you do need to include the empty password (the password with $0$ letters).
$endgroup$
– Alex Vong
Dec 13 '18 at 17:59
$begingroup$
I don't think there is a NULL letter. Lowercase letters are usually defined to be ${a, b, c, dots, z}$. However, I think you do need to include the empty password (the password with $0$ letters).
$endgroup$
– Alex Vong
Dec 13 '18 at 17:59
$begingroup$
@AlexVong There are plenty of reasons to consider a null character (especially in programming) and there are even some times where it might help with calculations. This happens to not be one of those cases, however that does not mean that it does not have its uses elsewhere.
$endgroup$
– JMoravitz
Dec 13 '18 at 18:03
$begingroup$
@AlexVong There are plenty of reasons to consider a null character (especially in programming) and there are even some times where it might help with calculations. This happens to not be one of those cases, however that does not mean that it does not have its uses elsewhere.
$endgroup$
– JMoravitz
Dec 13 '18 at 18:03
$begingroup$
@JMoravitz You're right. I should be more clear. What I meant to say is that the lowercase letters usually don't include the NULL character. Of course, we can have
'' in C!$endgroup$
– Alex Vong
Dec 13 '18 at 18:07
$begingroup$
@JMoravitz You're right. I should be more clear. What I meant to say is that the lowercase letters usually don't include the NULL character. Of course, we can have
'' in C!$endgroup$
– Alex Vong
Dec 13 '18 at 18:07
$begingroup$
It seems like your answer would allow blanks in the middle such as "ab bd ", is that what you want? If the blanks must be at the end, then it might be easier to add up the numbers of words of lengths, 1, 2, 3, ... , 9 and leave the blanks out of the computation. Also, I wouldn't imagine the empty string is a legitimate password in this context but that's just based on what I think "password" usually means in our world, not some technical math definition.
$endgroup$
– Ned
Dec 13 '18 at 18:07
$begingroup$
It seems like your answer would allow blanks in the middle such as "ab bd ", is that what you want? If the blanks must be at the end, then it might be easier to add up the numbers of words of lengths, 1, 2, 3, ... , 9 and leave the blanks out of the computation. Also, I wouldn't imagine the empty string is a legitimate password in this context but that's just based on what I think "password" usually means in our world, not some technical math definition.
$endgroup$
– Ned
Dec 13 '18 at 18:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would say it is $26 + 26^2 + 26^3 + cdots +26^9 = frac {26(26^9-1)}{25}$
On your arguement for the null letter. If we use _ to represent the null.
Is "a_b" a viable password?
The nulls must be the front or the back of the string. (and even then with a rule, all on the front or all on the back)
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
Your answer appears to follow the logic of: - Pick the first letter (from one of the 26 letters). -Pick the second letter (from one of the 27 letters where you include the null character). -Pick the third letter (from one of the 27 letters where you include the null character).
Letting + represent the null character, you are counting passwords which look not only like aabbccdde, aabb+++++ and a++++++++ but you are also counting passwords which look like a++a++b+b where you have letters occurring after the null characters.
If you consider the null character to be like an empty-string then you have accidentally overcounted aabb+++++ and a++a++b+b as "different" outcomes when they are in fact both simply aabb and should have been counted as the same. If you consider the null character to be like a "space" instead, then you are counting passwords which you should not be.
What you effectively counted was instead the number of passwords of length exactly $9$ where the first character must be a letter and the remaining characters may be a letter or a +.
For a corrected approach, count how many length $1$ passwords there are. Then count how many length $2$ passwords there are, then the length $3$ and so on up until length $9$ and add these together. (Depending on your interpretation, you might even consider a length 0 password)
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I would say it is $26 + 26^2 + 26^3 + cdots +26^9 = frac {26(26^9-1)}{25}$
On your arguement for the null letter. If we use _ to represent the null.
Is "a_b" a viable password?
The nulls must be the front or the back of the string. (and even then with a rule, all on the front or all on the back)
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
I would say it is $26 + 26^2 + 26^3 + cdots +26^9 = frac {26(26^9-1)}{25}$
On your arguement for the null letter. If we use _ to represent the null.
Is "a_b" a viable password?
The nulls must be the front or the back of the string. (and even then with a rule, all on the front or all on the back)
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
I would say it is $26 + 26^2 + 26^3 + cdots +26^9 = frac {26(26^9-1)}{25}$
On your arguement for the null letter. If we use _ to represent the null.
Is "a_b" a viable password?
The nulls must be the front or the back of the string. (and even then with a rule, all on the front or all on the back)
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
$endgroup$
I would say it is $26 + 26^2 + 26^3 + cdots +26^9 = frac {26(26^9-1)}{25}$
On your arguement for the null letter. If we use _ to represent the null.
Is "a_b" a viable password?
The nulls must be the front or the back of the string. (and even then with a rule, all on the front or all on the back)
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
answered Dec 13 '18 at 18:05
Doug MDoug M
44.9k31854
44.9k31854
add a comment |
add a comment |
$begingroup$
Your answer appears to follow the logic of: - Pick the first letter (from one of the 26 letters). -Pick the second letter (from one of the 27 letters where you include the null character). -Pick the third letter (from one of the 27 letters where you include the null character).
Letting + represent the null character, you are counting passwords which look not only like aabbccdde, aabb+++++ and a++++++++ but you are also counting passwords which look like a++a++b+b where you have letters occurring after the null characters.
If you consider the null character to be like an empty-string then you have accidentally overcounted aabb+++++ and a++a++b+b as "different" outcomes when they are in fact both simply aabb and should have been counted as the same. If you consider the null character to be like a "space" instead, then you are counting passwords which you should not be.
What you effectively counted was instead the number of passwords of length exactly $9$ where the first character must be a letter and the remaining characters may be a letter or a +.
For a corrected approach, count how many length $1$ passwords there are. Then count how many length $2$ passwords there are, then the length $3$ and so on up until length $9$ and add these together. (Depending on your interpretation, you might even consider a length 0 password)
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
$endgroup$
add a comment |
$begingroup$
Your answer appears to follow the logic of: - Pick the first letter (from one of the 26 letters). -Pick the second letter (from one of the 27 letters where you include the null character). -Pick the third letter (from one of the 27 letters where you include the null character).
Letting + represent the null character, you are counting passwords which look not only like aabbccdde, aabb+++++ and a++++++++ but you are also counting passwords which look like a++a++b+b where you have letters occurring after the null characters.
If you consider the null character to be like an empty-string then you have accidentally overcounted aabb+++++ and a++a++b+b as "different" outcomes when they are in fact both simply aabb and should have been counted as the same. If you consider the null character to be like a "space" instead, then you are counting passwords which you should not be.
What you effectively counted was instead the number of passwords of length exactly $9$ where the first character must be a letter and the remaining characters may be a letter or a +.
For a corrected approach, count how many length $1$ passwords there are. Then count how many length $2$ passwords there are, then the length $3$ and so on up until length $9$ and add these together. (Depending on your interpretation, you might even consider a length 0 password)
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
$endgroup$
add a comment |
$begingroup$
Your answer appears to follow the logic of: - Pick the first letter (from one of the 26 letters). -Pick the second letter (from one of the 27 letters where you include the null character). -Pick the third letter (from one of the 27 letters where you include the null character).
Letting + represent the null character, you are counting passwords which look not only like aabbccdde, aabb+++++ and a++++++++ but you are also counting passwords which look like a++a++b+b where you have letters occurring after the null characters.
If you consider the null character to be like an empty-string then you have accidentally overcounted aabb+++++ and a++a++b+b as "different" outcomes when they are in fact both simply aabb and should have been counted as the same. If you consider the null character to be like a "space" instead, then you are counting passwords which you should not be.
What you effectively counted was instead the number of passwords of length exactly $9$ where the first character must be a letter and the remaining characters may be a letter or a +.
For a corrected approach, count how many length $1$ passwords there are. Then count how many length $2$ passwords there are, then the length $3$ and so on up until length $9$ and add these together. (Depending on your interpretation, you might even consider a length 0 password)
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
$endgroup$
Your answer appears to follow the logic of: - Pick the first letter (from one of the 26 letters). -Pick the second letter (from one of the 27 letters where you include the null character). -Pick the third letter (from one of the 27 letters where you include the null character).
Letting + represent the null character, you are counting passwords which look not only like aabbccdde, aabb+++++ and a++++++++ but you are also counting passwords which look like a++a++b+b where you have letters occurring after the null characters.
If you consider the null character to be like an empty-string then you have accidentally overcounted aabb+++++ and a++a++b+b as "different" outcomes when they are in fact both simply aabb and should have been counted as the same. If you consider the null character to be like a "space" instead, then you are counting passwords which you should not be.
What you effectively counted was instead the number of passwords of length exactly $9$ where the first character must be a letter and the remaining characters may be a letter or a +.
For a corrected approach, count how many length $1$ passwords there are. Then count how many length $2$ passwords there are, then the length $3$ and so on up until length $9$ and add these together. (Depending on your interpretation, you might even consider a length 0 password)
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
answered Dec 13 '18 at 17:58
JMoravitzJMoravitz
47.5k33886
47.5k33886
add a comment |
add a comment |
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1
$begingroup$
Welcome to stackexchange. Please edit the question to show us how you arrived at that answer. Then we may be able to help you. Perhaps start by answering the question with $2$ or $3$ instead of $9$ and an alphabet with fewer letters to see the pattern. Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:50
$begingroup$
I don't think there is a NULL letter. Lowercase letters are usually defined to be ${a, b, c, dots, z}$. However, I think you do need to include the empty password (the password with $0$ letters).
$endgroup$
– Alex Vong
Dec 13 '18 at 17:59
$begingroup$
@AlexVong There are plenty of reasons to consider a null character (especially in programming) and there are even some times where it might help with calculations. This happens to not be one of those cases, however that does not mean that it does not have its uses elsewhere.
$endgroup$
– JMoravitz
Dec 13 '18 at 18:03
$begingroup$
@JMoravitz You're right. I should be more clear. What I meant to say is that the lowercase letters usually don't include the NULL character. Of course, we can have
''in C!$endgroup$
– Alex Vong
Dec 13 '18 at 18:07
$begingroup$
It seems like your answer would allow blanks in the middle such as "ab bd ", is that what you want? If the blanks must be at the end, then it might be easier to add up the numbers of words of lengths, 1, 2, 3, ... , 9 and leave the blanks out of the computation. Also, I wouldn't imagine the empty string is a legitimate password in this context but that's just based on what I think "password" usually means in our world, not some technical math definition.
$endgroup$
– Ned
Dec 13 '18 at 18:07