Mixing
Number of ways to form a password of length 8 9 and 10 with restrictions
$begingroup$
Consider creating a password, and Its length is between $8$ and $10$ characters. Each character is either lowercase letters without accents, uppercase letters without accents, digits from 0 to 9 or the special characters !?#&@%*~$_. The password contains at least one character of at least three of the four types mentioned. In how many ways can a password be chosen to meet all those requirements?
Here is my attempt: There are $10*26*26+ 10*26*26+26*10*10+26*10*10$ ways to pick 3 characters one from three of the four types. There are $binom{8}{3}+binom{9}{3}+binom{10}{3}$ to choose three positions for the length 8, 9 and 10 strings. So then:
$$(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{8}{3}*frac{72!}{(72-5)!}+ (10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{9}{3}*frac{72!}{(72-6)!} +(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{10}{3}*frac{72!}{(72-7)!}$$
But this clearly has overlaps, so I overcounted. How can I solve this problem?
combinatorics
edited Feb 1 at 7:03
jvdhooft
5,65961641
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
$endgroup$
add a comment |
$begingroup$
Consider creating a password, and Its length is between $8$ and $10$ characters. Each character is either lowercase letters without accents, uppercase letters without accents, digits from 0 to 9 or the special characters !?#&@%*~$_. The password contains at least one character of at least three of the four types mentioned. In how many ways can a password be chosen to meet all those requirements?
Here is my attempt: There are $10*26*26+ 10*26*26+26*10*10+26*10*10$ ways to pick 3 characters one from three of the four types. There are $binom{8}{3}+binom{9}{3}+binom{10}{3}$ to choose three positions for the length 8, 9 and 10 strings. So then:
$$(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{8}{3}*frac{72!}{(72-5)!}+ (10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{9}{3}*frac{72!}{(72-6)!} +(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{10}{3}*frac{72!}{(72-7)!}$$
But this clearly has overlaps, so I overcounted. How can I solve this problem?
combinatorics
edited Feb 1 at 7:03
jvdhooft
5,65961641
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
$endgroup$
add a comment |
$begingroup$
Consider creating a password, and Its length is between $8$ and $10$ characters. Each character is either lowercase letters without accents, uppercase letters without accents, digits from 0 to 9 or the special characters !?#&@%*~$_. The password contains at least one character of at least three of the four types mentioned. In how many ways can a password be chosen to meet all those requirements?
Here is my attempt: There are $10*26*26+ 10*26*26+26*10*10+26*10*10$ ways to pick 3 characters one from three of the four types. There are $binom{8}{3}+binom{9}{3}+binom{10}{3}$ to choose three positions for the length 8, 9 and 10 strings. So then:
$$(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{8}{3}*frac{72!}{(72-5)!}+ (10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{9}{3}*frac{72!}{(72-6)!} +(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{10}{3}*frac{72!}{(72-7)!}$$
But this clearly has overlaps, so I overcounted. How can I solve this problem?
combinatorics
edited Feb 1 at 7:03
jvdhooft
5,65961641
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
$endgroup$
Consider creating a password, and Its length is between $8$ and $10$ characters. Each character is either lowercase letters without accents, uppercase letters without accents, digits from 0 to 9 or the special characters !?#&@%*~$_. The password contains at least one character of at least three of the four types mentioned. In how many ways can a password be chosen to meet all those requirements?
Here is my attempt: There are $10*26*26+ 10*26*26+26*10*10+26*10*10$ ways to pick 3 characters one from three of the four types. There are $binom{8}{3}+binom{9}{3}+binom{10}{3}$ to choose three positions for the length 8, 9 and 10 strings. So then:
$$(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{8}{3}*frac{72!}{(72-5)!}+ (10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{9}{3}*frac{72!}{(72-6)!} +(10*26*26+ 10*26*26+26*10*10+26*10*10)*binom{10}{3}*frac{72!}{(72-7)!}$$
But this clearly has overlaps, so I overcounted. How can I solve this problem?
combinatorics
combinatorics
edited Feb 1 at 7:03
jvdhooft
5,65961641
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
edited Feb 1 at 7:03
jvdhooft
5,65961641
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
edited Feb 1 at 7:03
jvdhooft
5,65961641
edited Feb 1 at 7:03
jvdhooft
5,65961641
edited Feb 1 at 7:03
jvdhooft
5,65961641
5,65961641
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
asked Feb 1 at 5:08
mathpadawanmathpadawan
1,940422
1,940422
add a comment |
add a comment |
1 Answer
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$begingroup$
We can use subtraction to solve this problem. If we would have one character for each group only, the number of possible combinations of length $n$ would be:
$$4^n - {4 choose 2} (2^n - 2) - {4 choose 1} 1^n$$
This comes down to considering all possible combinations with four groups of characters, subtracting all combinations with two groups or less. We can use a similar approach when multiple characters are allowed. For the number of combinations $f(n)$ of length $n$, we find:
$$f(n) = 72^n - [(52^n - 2 cdot 26^n) + 4 cdot (36^n - 26^n -10^n) + (20^n - 2 cdot 10^n)] - (2 cdot 26^n + 2 cdot 10^n)$$
The total number of valid passwords thus equals:
$$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
We can use subtraction to solve this problem. If we would have one character for each group only, the number of possible combinations of length $n$ would be:
$$4^n - {4 choose 2} (2^n - 2) - {4 choose 1} 1^n$$
This comes down to considering all possible combinations with four groups of characters, subtracting all combinations with two groups or less. We can use a similar approach when multiple characters are allowed. For the number of combinations $f(n)$ of length $n$, we find:
$$f(n) = 72^n - [(52^n - 2 cdot 26^n) + 4 cdot (36^n - 26^n -10^n) + (20^n - 2 cdot 10^n)] - (2 cdot 26^n + 2 cdot 10^n)$$
The total number of valid passwords thus equals:
$$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
We can use subtraction to solve this problem. If we would have one character for each group only, the number of possible combinations of length $n$ would be:
$$4^n - {4 choose 2} (2^n - 2) - {4 choose 1} 1^n$$
This comes down to considering all possible combinations with four groups of characters, subtracting all combinations with two groups or less. We can use a similar approach when multiple characters are allowed. For the number of combinations $f(n)$ of length $n$, we find:
$$f(n) = 72^n - [(52^n - 2 cdot 26^n) + 4 cdot (36^n - 26^n -10^n) + (20^n - 2 cdot 10^n)] - (2 cdot 26^n + 2 cdot 10^n)$$
The total number of valid passwords thus equals:
$$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
We can use subtraction to solve this problem. If we would have one character for each group only, the number of possible combinations of length $n$ would be:
$$4^n - {4 choose 2} (2^n - 2) - {4 choose 1} 1^n$$
This comes down to considering all possible combinations with four groups of characters, subtracting all combinations with two groups or less. We can use a similar approach when multiple characters are allowed. For the number of combinations $f(n)$ of length $n$, we find:
$$f(n) = 72^n - [(52^n - 2 cdot 26^n) + 4 cdot (36^n - 26^n -10^n) + (20^n - 2 cdot 10^n)] - (2 cdot 26^n + 2 cdot 10^n)$$
The total number of valid passwords thus equals:
$$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
$endgroup$
We can use subtraction to solve this problem. If we would have one character for each group only, the number of possible combinations of length $n$ would be:
$$4^n - {4 choose 2} (2^n - 2) - {4 choose 1} 1^n$$
This comes down to considering all possible combinations with four groups of characters, subtracting all combinations with two groups or less. We can use a similar approach when multiple characters are allowed. For the number of combinations $f(n)$ of length $n$, we find:
$$f(n) = 72^n - [(52^n - 2 cdot 26^n) + 4 cdot (36^n - 26^n -10^n) + (20^n - 2 cdot 10^n)] - (2 cdot 26^n + 2 cdot 10^n)$$
The total number of valid passwords thus equals:
$$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
edited Feb 1 at 7:29
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 7:01
jvdhooftjvdhooft
5,65961641
5,65961641
add a comment |
add a comment |
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