# Permutations and Combinations problem

• Nov 1st 2009, 06:05 PM
dancecubed
Permutations and Combinations problem
This one has really been troubling me. The question is:

A computer program requires a password of at least 6 but no more than 8 characters. All letters (of the English alphabet, not case sensitive) and digits (0 to 9) may be used. The password must contain at least one digit and one letter. How many different passwords are possible?

I used the indirect method for each case and got down to:

Case 1: 6 character password

36^6 - (26^6 + 10^6)

Case 2: 7 character password

36^7 - (26^7 + 10^7)

Case 3: 8 character password

36^8 - (26^8 + 10^8)

And then I added all the cases together. Is that correct?
• Nov 2nd 2009, 08:14 AM
Hello dancecubed

Welcome to Math Help Forum!
Quote:

Originally Posted by dancecubed
This one has really been troubling me. The question is:

A computer program requires a password of at least 6 but no more than 8 characters. All letters (of the English alphabet, not case sensitive) and digits (0 to 9) may be used. The password must contain at least one digit and one letter. How many different passwords are possible?

I used the indirect method for each case and got down to:

Case 1: 6 character password

36^6 - (26^6 + 10^6)

Case 2: 7 character password

36^7 - (26^7 + 10^7)

Case 3: 8 character password

36^8 - (26^8 + 10^8)

And then I added all the cases together. Is that correct?

I think it is quite correct. For instance:

In Case 1 (6-character passwords):
There are \$\displaystyle 36^6\$ passwords altogether if there is no restriction, since there are \$\displaystyle 36\$ choices for each of \$\displaystyle 6\$ positions, repetitions being allowed.

Of these, there are \$\displaystyle 26^6\$ that will contain no digits, and \$\displaystyle 10^6\$ that will contain no letters.

There are therefore \$\displaystyle 36^6-(26^6+10^6)\$ that each contain at least one letter and one digit.
Similarly for the 7- and 8- character passwords.