Mixing
Combinatorics - counting allowed passwords
$begingroup$
I'm struggling with the following problem just because I get from the calculations numbers too big.
To register for a site you can choose a password consisting of a minimum of 4 to a maximum of 8 characters chosen from 26 with a distinction between upper and lower case. How many passwords are available?
My attempt. Thinking with a string of length 4, I have a set of 52 characters to build a string. I only have to pay attention to the order of the characters as this distinguishes one password from another. However, I can repeat the same character several times. So for the first character, I have 52 choices, 52 for the second one etc. In total 52^4. I applied the same reasoning for passwords of length 5,6,7,8 and at the end, I just add these results that I got separately to know all the possible passwords.
Is my reasoning correct?
combinatorics
asked Jan 18 at 17:52
PCNFPCNF
1338
$endgroup$
add a comment |
$begingroup$
I'm struggling with the following problem just because I get from the calculations numbers too big.
To register for a site you can choose a password consisting of a minimum of 4 to a maximum of 8 characters chosen from 26 with a distinction between upper and lower case. How many passwords are available?
My attempt. Thinking with a string of length 4, I have a set of 52 characters to build a string. I only have to pay attention to the order of the characters as this distinguishes one password from another. However, I can repeat the same character several times. So for the first character, I have 52 choices, 52 for the second one etc. In total 52^4. I applied the same reasoning for passwords of length 5,6,7,8 and at the end, I just add these results that I got separately to know all the possible passwords.
Is my reasoning correct?
combinatorics
asked Jan 18 at 17:52
PCNFPCNF
1338
$endgroup$
$begingroup$
You are correct. The answer should be 54,507,958,359,296.
$endgroup$
– user3482749
Jan 18 at 17:56
add a comment |
$begingroup$
I'm struggling with the following problem just because I get from the calculations numbers too big.
To register for a site you can choose a password consisting of a minimum of 4 to a maximum of 8 characters chosen from 26 with a distinction between upper and lower case. How many passwords are available?
My attempt. Thinking with a string of length 4, I have a set of 52 characters to build a string. I only have to pay attention to the order of the characters as this distinguishes one password from another. However, I can repeat the same character several times. So for the first character, I have 52 choices, 52 for the second one etc. In total 52^4. I applied the same reasoning for passwords of length 5,6,7,8 and at the end, I just add these results that I got separately to know all the possible passwords.
Is my reasoning correct?
combinatorics
asked Jan 18 at 17:52
PCNFPCNF
1338
$endgroup$
I'm struggling with the following problem just because I get from the calculations numbers too big.
To register for a site you can choose a password consisting of a minimum of 4 to a maximum of 8 characters chosen from 26 with a distinction between upper and lower case. How many passwords are available?
My attempt. Thinking with a string of length 4, I have a set of 52 characters to build a string. I only have to pay attention to the order of the characters as this distinguishes one password from another. However, I can repeat the same character several times. So for the first character, I have 52 choices, 52 for the second one etc. In total 52^4. I applied the same reasoning for passwords of length 5,6,7,8 and at the end, I just add these results that I got separately to know all the possible passwords.
Is my reasoning correct?
combinatorics
combinatorics
asked Jan 18 at 17:52
PCNFPCNF
1338
asked Jan 18 at 17:52
PCNFPCNF
1338
asked Jan 18 at 17:52
PCNFPCNF
1338
asked Jan 18 at 17:52
PCNFPCNF
1338
asked Jan 18 at 17:52
PCNFPCNF
1338
1338
$begingroup$
You are correct. The answer should be 54,507,958,359,296.
$endgroup$
– user3482749
Jan 18 at 17:56
add a comment |
$begingroup$
You are correct. The answer should be 54,507,958,359,296.
$endgroup$
– user3482749
Jan 18 at 17:56
$begingroup$
You are correct. The answer should be 54,507,958,359,296.
$endgroup$
– user3482749
Jan 18 at 17:56
$begingroup$
You are correct. The answer should be 54,507,958,359,296.
$endgroup$
– user3482749
Jan 18 at 17:56
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Yes, your reasoning is correct. As there are $26times 2 = 52$ characters for each slot, and each length of password from $4$ to $8$ is allowed, this results in a total of:
$$52^4+52^5+52^6+52^7+52^8$$
In a classroom environment, this answer is perfectly acceptable without simplifying to a single whole number result. In fact, I would argue that this answer is preferred since seeing it in this form clearly implies what some of the thought process behind the answer is and is easily verifiable as being correct.
If you insist on having the answer be simplified fully, you can use more powerful calculators or programming languages than the ones you were previously using and get an answer of 54507958359296.
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Yes, your reasoning is correct. As there are $26times 2 = 52$ characters for each slot, and each length of password from $4$ to $8$ is allowed, this results in a total of:
$$52^4+52^5+52^6+52^7+52^8$$
In a classroom environment, this answer is perfectly acceptable without simplifying to a single whole number result. In fact, I would argue that this answer is preferred since seeing it in this form clearly implies what some of the thought process behind the answer is and is easily verifiable as being correct.
If you insist on having the answer be simplified fully, you can use more powerful calculators or programming languages than the ones you were previously using and get an answer of 54507958359296.
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
$endgroup$
add a comment |
$begingroup$
Yes, your reasoning is correct. As there are $26times 2 = 52$ characters for each slot, and each length of password from $4$ to $8$ is allowed, this results in a total of:
$$52^4+52^5+52^6+52^7+52^8$$
In a classroom environment, this answer is perfectly acceptable without simplifying to a single whole number result. In fact, I would argue that this answer is preferred since seeing it in this form clearly implies what some of the thought process behind the answer is and is easily verifiable as being correct.
If you insist on having the answer be simplified fully, you can use more powerful calculators or programming languages than the ones you were previously using and get an answer of 54507958359296.
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
$endgroup$
add a comment |
$begingroup$
Yes, your reasoning is correct. As there are $26times 2 = 52$ characters for each slot, and each length of password from $4$ to $8$ is allowed, this results in a total of:
$$52^4+52^5+52^6+52^7+52^8$$
In a classroom environment, this answer is perfectly acceptable without simplifying to a single whole number result. In fact, I would argue that this answer is preferred since seeing it in this form clearly implies what some of the thought process behind the answer is and is easily verifiable as being correct.
If you insist on having the answer be simplified fully, you can use more powerful calculators or programming languages than the ones you were previously using and get an answer of 54507958359296.
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
$endgroup$
Yes, your reasoning is correct. As there are $26times 2 = 52$ characters for each slot, and each length of password from $4$ to $8$ is allowed, this results in a total of:
$$52^4+52^5+52^6+52^7+52^8$$
In a classroom environment, this answer is perfectly acceptable without simplifying to a single whole number result. In fact, I would argue that this answer is preferred since seeing it in this form clearly implies what some of the thought process behind the answer is and is easily verifiable as being correct.
If you insist on having the answer be simplified fully, you can use more powerful calculators or programming languages than the ones you were previously using and get an answer of 54507958359296.
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
answered Jan 18 at 18:11
JMoravitzJMoravitz
48k33886
48k33886
add a comment |
add a comment |
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$begingroup$
You are correct. The answer should be 54,507,958,359,296.
$endgroup$
– user3482749
Jan 18 at 17:56