How many 4 digit pins on set {0-9}
$begingroup$
A password can be any 4 digit {0...9}.
1.)How many possible passwords are there?
for this I did $10^4 = 10,000$
2.) How many possible passwords with no repeated digits?
$10*9*8*7 = 5040$
3.) How many possible passwords are there with repeated digits?
For this do I just subtract 10,000-5040?
Edit: typos. Thanks everyone.
combinatorics permutations
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
$endgroup$
add a comment |
$begingroup$
A password can be any 4 digit {0...9}.
1.)How many possible passwords are there?
for this I did $10^4 = 10,000$
2.) How many possible passwords with no repeated digits?
$10*9*8*7 = 5040$
3.) How many possible passwords are there with repeated digits?
For this do I just subtract 10,000-5040?
Edit: typos. Thanks everyone.
combinatorics permutations
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
$endgroup$
2
$begingroup$
I think you mean 1 should be $10^{4}$..
$endgroup$
– Mattos
May 24 '16 at 2:06
1
$begingroup$
You are correct. Repeated digits and no repeated digits are mutually exclusive and span the entire set of passwords so what you did works.
$endgroup$
– lordoftheshadows
May 24 '16 at 2:09
1
$begingroup$
Typo. $10^4$. Yes. Pins either have repeated digits or they don't. So REPEAT +NON =ALL so REPEAT = ALL - NON. That's ligical, isn't it? (And yes, other than the typi, you did 1 and 2 perfectly correctly.)
$endgroup$
– fleablood
May 24 '16 at 2:10
add a comment |
$begingroup$
A password can be any 4 digit {0...9}.
1.)How many possible passwords are there?
for this I did $10^4 = 10,000$
2.) How many possible passwords with no repeated digits?
$10*9*8*7 = 5040$
3.) How many possible passwords are there with repeated digits?
For this do I just subtract 10,000-5040?
Edit: typos. Thanks everyone.
combinatorics permutations
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
$endgroup$
A password can be any 4 digit {0...9}.
1.)How many possible passwords are there?
for this I did $10^4 = 10,000$
2.) How many possible passwords with no repeated digits?
$10*9*8*7 = 5040$
3.) How many possible passwords are there with repeated digits?
For this do I just subtract 10,000-5040?
Edit: typos. Thanks everyone.
combinatorics permutations
combinatorics permutations
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
edited May 24 '16 at 9:47
N. F. Taussig
44.4k93357
44.4k93357
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
asked May 24 '16 at 2:02
OrangePineappleOrangePineapple
7229
7229
2
$begingroup$
I think you mean 1 should be $10^{4}$..
$endgroup$
– Mattos
May 24 '16 at 2:06
1
$begingroup$
You are correct. Repeated digits and no repeated digits are mutually exclusive and span the entire set of passwords so what you did works.
$endgroup$
– lordoftheshadows
May 24 '16 at 2:09
1
$begingroup$
Typo. $10^4$. Yes. Pins either have repeated digits or they don't. So REPEAT +NON =ALL so REPEAT = ALL - NON. That's ligical, isn't it? (And yes, other than the typi, you did 1 and 2 perfectly correctly.)
$endgroup$
– fleablood
May 24 '16 at 2:10
add a comment |
2
$begingroup$
I think you mean 1 should be $10^{4}$..
$endgroup$
– Mattos
May 24 '16 at 2:06
1
$begingroup$
You are correct. Repeated digits and no repeated digits are mutually exclusive and span the entire set of passwords so what you did works.
$endgroup$
– lordoftheshadows
May 24 '16 at 2:09
1
$begingroup$
Typo. $10^4$. Yes. Pins either have repeated digits or they don't. So REPEAT +NON =ALL so REPEAT = ALL - NON. That's ligical, isn't it? (And yes, other than the typi, you did 1 and 2 perfectly correctly.)
$endgroup$
– fleablood
May 24 '16 at 2:10
2
2
$begingroup$
I think you mean 1 should be $10^{4}$..
$endgroup$
– Mattos
May 24 '16 at 2:06
$begingroup$
I think you mean 1 should be $10^{4}$..
$endgroup$
– Mattos
May 24 '16 at 2:06
1
1
$begingroup$
You are correct. Repeated digits and no repeated digits are mutually exclusive and span the entire set of passwords so what you did works.
$endgroup$
– lordoftheshadows
May 24 '16 at 2:09
$begingroup$
You are correct. Repeated digits and no repeated digits are mutually exclusive and span the entire set of passwords so what you did works.
$endgroup$
– lordoftheshadows
May 24 '16 at 2:09
1
1
$begingroup$
Typo. $10^4$. Yes. Pins either have repeated digits or they don't. So REPEAT +NON =ALL so REPEAT = ALL - NON. That's ligical, isn't it? (And yes, other than the typi, you did 1 and 2 perfectly correctly.)
$endgroup$
– fleablood
May 24 '16 at 2:10
$begingroup$
Typo. $10^4$. Yes. Pins either have repeated digits or they don't. So REPEAT +NON =ALL so REPEAT = ALL - NON. That's ligical, isn't it? (And yes, other than the typi, you did 1 and 2 perfectly correctly.)
$endgroup$
– fleablood
May 24 '16 at 2:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
community wiki answer so the question can be closed
Your solutions to the first two questions are correct.
Yes, to find the number of passwords with repeated digits, you subtract the number of passwords with no repeated digits from the total, so there are $10000 - 5040 = 4960$, as you found.
$endgroup$
add a comment |
$begingroup$
For question 2, if you don't consider $0$ as a possible digit for the first number (because then it would be a 3 digit number, rather than a 4 digit number), then the answer is $9*9*8*7 = 4536$.
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
$endgroup$
add a comment |
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2 Answers
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$begingroup$
community wiki answer so the question can be closed
Your solutions to the first two questions are correct.
Yes, to find the number of passwords with repeated digits, you subtract the number of passwords with no repeated digits from the total, so there are $10000 - 5040 = 4960$, as you found.
$endgroup$
add a comment |
$begingroup$
community wiki answer so the question can be closed
Your solutions to the first two questions are correct.
Yes, to find the number of passwords with repeated digits, you subtract the number of passwords with no repeated digits from the total, so there are $10000 - 5040 = 4960$, as you found.
$endgroup$
add a comment |
$begingroup$
community wiki answer so the question can be closed
Your solutions to the first two questions are correct.
Yes, to find the number of passwords with repeated digits, you subtract the number of passwords with no repeated digits from the total, so there are $10000 - 5040 = 4960$, as you found.
$endgroup$
community wiki answer so the question can be closed
Your solutions to the first two questions are correct.
Yes, to find the number of passwords with repeated digits, you subtract the number of passwords with no repeated digits from the total, so there are $10000 - 5040 = 4960$, as you found.
edited Apr 22 '18 at 9:57
community wiki
3 revs, 2 users 88%
N. F. Taussig
add a comment |
add a comment |
$begingroup$
For question 2, if you don't consider $0$ as a possible digit for the first number (because then it would be a 3 digit number, rather than a 4 digit number), then the answer is $9*9*8*7 = 4536$.
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
$endgroup$
add a comment |
$begingroup$
For question 2, if you don't consider $0$ as a possible digit for the first number (because then it would be a 3 digit number, rather than a 4 digit number), then the answer is $9*9*8*7 = 4536$.
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
$endgroup$
add a comment |
$begingroup$
For question 2, if you don't consider $0$ as a possible digit for the first number (because then it would be a 3 digit number, rather than a 4 digit number), then the answer is $9*9*8*7 = 4536$.
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
$endgroup$
For question 2, if you don't consider $0$ as a possible digit for the first number (because then it would be a 3 digit number, rather than a 4 digit number), then the answer is $9*9*8*7 = 4536$.
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
answered Dec 28 '18 at 19:02
Isopycnal OscillationIsopycnal Oscillation
1155
1155
add a comment |
add a comment |
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2
$begingroup$
I think you mean 1 should be $10^{4}$..
$endgroup$
– Mattos
May 24 '16 at 2:06
1
$begingroup$
You are correct. Repeated digits and no repeated digits are mutually exclusive and span the entire set of passwords so what you did works.
$endgroup$
– lordoftheshadows
May 24 '16 at 2:09
1
$begingroup$
Typo. $10^4$. Yes. Pins either have repeated digits or they don't. So REPEAT +NON =ALL so REPEAT = ALL - NON. That's ligical, isn't it? (And yes, other than the typi, you did 1 and 2 perfectly correctly.)
$endgroup$
– fleablood
May 24 '16 at 2:10