Counting problem

  • Thread starter Miike012
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Im trying to find all combinations of P6. Book solution in paint doc.

My solution: Please tell me where I am going wrong.

P6: Password of 6 characters
1. Each password must contain at least one digit,
2. Each character of password can be a digit or uppercase letter.

Let P61 be defined as follows. Ci is the ith character of P61, i = 1,...,6

Let C1 be a digit then the following characters can be a digit or uppercase letter.
C1 has 10 choices and Ci has 36 choices for i = 2,..,6.

Therefore the password defined by P61, which was defined by restricting C1 to be a digit, has a total of 10*365 choices.

I will do the same for P6i, i = 2,...6, where the ith character is a digit.

All passwords P6i i = 1,2,3,4,5,6 will look like the following:
Let D represent the character that is a digit and DL represent the character that is a digit or uppercase letter.

P61 P62 ... P66
1.D 1.DL 1.DL
2.DL 2.D 2.DL
3.DL 3.DL 3.DL
4.DL 4.DL 4.DL
5.DL 5.DL 5.DL
6.DL 6.DL 6.D


Hence you can see that there are 6 total different "sub catagories" of P6 and there must be 10*365 choices per sub catagoy.
Therefore total choices for P6
Ʃ(Number of choices for P6i) i = 1,...,6
= Ʃ(10*365) i = 1,...,6
= 6*10*365



What am I counting extra of?
 

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  • #2
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  • #3
verty
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A password like 11XXXX will appear in group 1 and group 2.
 
  • #4
haruspex
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How many passwords satisfy (2). How many of those do not satisfy (1)?
 

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