Number of ways to form a password of length 8 9 and 10 with restrictions












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$begingroup$



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







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    1












    $begingroup$



    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      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







      share|cite|improve this question











      $endgroup$





      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






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      share|cite|improve this question













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      share|cite|improve this question








      edited Feb 1 at 7:03









      jvdhooft

      5,65961641




      5,65961641










      asked Feb 1 at 5:08









      mathpadawanmathpadawan

      1,940422




      1,940422






















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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:



          $$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$






          share|cite|improve this 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:



            $$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              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:



              $$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                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:



                $$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$






                share|cite|improve this answer











                $endgroup$



                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:



                $$f(8) + f(9) + f(10) approx 3.635 cdot 10^{18}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 1 at 7:29

























                answered Feb 1 at 7:01









                jvdhooftjvdhooft

                5,65961641




                5,65961641






























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