counting pasword character when repeated allowed [closed]
$begingroup$
How many six character password can be made using only A, B,C,D,E,F, 1, 2, 3 , 4, 5, 6 if
i. No character is reused
ii. character can be repeated as long as they are not adjacent
combinatorics
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
$endgroup$
closed as off-topic by Namaste, John Douma, Saad, Tianlalu, Leucippus Dec 24 '18 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
How many six character password can be made using only A, B,C,D,E,F, 1, 2, 3 , 4, 5, 6 if
i. No character is reused
ii. character can be repeated as long as they are not adjacent
combinatorics
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
$endgroup$
closed as off-topic by Namaste, John Douma, Saad, Tianlalu, Leucippus Dec 24 '18 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, John Douma, Saad, Tianlalu, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How many six character password can be made using only A, B,C,D,E,F, 1, 2, 3 , 4, 5, 6 if
i. No character is reused
ii. character can be repeated as long as they are not adjacent
combinatorics
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
$endgroup$
How many six character password can be made using only A, B,C,D,E,F, 1, 2, 3 , 4, 5, 6 if
i. No character is reused
ii. character can be repeated as long as they are not adjacent
combinatorics
combinatorics
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
edited Dec 24 '18 at 5:17
Hajer Bagdady
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
asked Dec 23 '18 at 18:01
Hajer BagdadyHajer Bagdady
41
41
closed as off-topic by Namaste, John Douma, Saad, Tianlalu, Leucippus Dec 24 '18 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, John Douma, Saad, Tianlalu, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, John Douma, Saad, Tianlalu, Leucippus Dec 24 '18 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, John Douma, Saad, Tianlalu, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
1
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votes
$begingroup$
You have $12$ choices for the first character.
You then have $11$ choices for the second character (anything except whatever you chose for the first character).
Similarly, $11$ choices for the third character (anything except whatever you chose for the second character).
And also $11$ choices for the fourth, fifth and sixth character.
Hence, there is a total of $$12 times 11 times 11 times 11 times 11 times 11 = 1932612$$ possible passwords.
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have $12$ choices for the first character.
You then have $11$ choices for the second character (anything except whatever you chose for the first character).
Similarly, $11$ choices for the third character (anything except whatever you chose for the second character).
And also $11$ choices for the fourth, fifth and sixth character.
Hence, there is a total of $$12 times 11 times 11 times 11 times 11 times 11 = 1932612$$ possible passwords.
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
$endgroup$
add a comment |
$begingroup$
You have $12$ choices for the first character.
You then have $11$ choices for the second character (anything except whatever you chose for the first character).
Similarly, $11$ choices for the third character (anything except whatever you chose for the second character).
And also $11$ choices for the fourth, fifth and sixth character.
Hence, there is a total of $$12 times 11 times 11 times 11 times 11 times 11 = 1932612$$ possible passwords.
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
$endgroup$
add a comment |
$begingroup$
You have $12$ choices for the first character.
You then have $11$ choices for the second character (anything except whatever you chose for the first character).
Similarly, $11$ choices for the third character (anything except whatever you chose for the second character).
And also $11$ choices for the fourth, fifth and sixth character.
Hence, there is a total of $$12 times 11 times 11 times 11 times 11 times 11 = 1932612$$ possible passwords.
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
$endgroup$
You have $12$ choices for the first character.
You then have $11$ choices for the second character (anything except whatever you chose for the first character).
Similarly, $11$ choices for the third character (anything except whatever you chose for the second character).
And also $11$ choices for the fourth, fifth and sixth character.
Hence, there is a total of $$12 times 11 times 11 times 11 times 11 times 11 = 1932612$$ possible passwords.
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
answered Dec 23 '18 at 18:09
glowstonetreesglowstonetrees
2,375418
2,375418
add a comment |
add a comment |
