Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > University Math Help > Discrete Mathematics, Set Theory and Logic
Reply
 
Thread Tools Display Modes
  #1  
Old November 1st, 2009, 07:05 PM
Newbie
 
Join Date: Oct 2009
Posts: 5
Country:
Thanks: 2
Thanked 0 Times in 0 Posts
dancecubed is on a distinguished road
Default 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?
Reply With Quote
Advertisement
 
  #2  
Old November 2nd, 2009, 09:14 AM
Grandad's Avatar
MHF Contributor

 
Join Date: Dec 2008
Location: South Coast of England
Posts: 2,295
Country:
Thanks: 158
Thanked 1,283 Times in 1,124 Posts
Grandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud ofGrandad has much to be proud of
Default

Hello dancecubed

Welcome to Math Help Forum!
Quote:
Originally Posted by dancecubed View Post
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 36^6 passwords altogether if there is no restriction, since there are 36 choices for each of 6 positions, repetitions being allowed.

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

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

Grandad
Reply With Quote
The following users thank Grandad for this useful post:
Donate to MHF
  #3  
Old November 3rd, 2009, 09:17 PM
Newbie
 
Join Date: Oct 2009
Posts: 5
Country:
Thanks: 2
Thanked 0 Times in 0 Posts
dancecubed is on a distinguished road
Default

Thank you very much! I've handed in my answers and now only time will tell
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 01:05 AM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2010, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.