BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Counting and Probability HELP!!!

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Fri Jan 28, 2011 9:52 pm

Counting and Probability HELP!!!

by shriti » Fri Feb 18, 2011 9:56 am
A popular website requires users to create a password consisting of digits only. If no digit may be repeated and each password must be at least 9 digits long, how many passwords are possible?

a. 9!+10!
b) 2!*10!
c) 9!*10!
d) 19!
e) 20!
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Jan 17, 2010 2:33 am

by Pilot » Fri Feb 18, 2011 10:27 am
The answer : B
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 127
Joined: Sat Oct 02, 2010 8:55 am
Thanked: 17 times
Followed by:1 members

by XLogic » Fri Feb 18, 2011 11:13 am
Please explain why the answer is 9! * 2!
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Jan 17, 2010 2:33 am

by Pilot » Fri Feb 18, 2011 11:49 am
Step 1 .The number of 10 digit password is 10!.

Step 2. We count 9 digit passwords. First we choose 9 digit sets out of 10 numbers. number of sets is equal to 10C9=10.

Then, from each set we can create 9! passwords. Hence the number of 9 digit passwords is 10*9!=10!.

Totally : 10!+10!=2*10!
Top
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Fri Feb 18, 2011 12:31 pm
Hi pilot, it does not really matter in this problem. However, if the problem's condition were must be at least 8 digits we should have used an ordered arrangement (permutation). Hence I propose instead of 10C9 to use 10P9 (P, permutation)
Pilot wrote:Step 1 .The number of 10 digit password is 10!.

Step 2. We count 9 digit passwords. First we choose 9 digit sets out of 10 numbers. number of sets is equal to 10C9=10.

Then, from each set we can create 9! passwords. Hence the number of 9 digit passwords is 10*9!=10!.

Totally : 10!+10!=2*10!
A popular website requires users to create a password consisting of digits only. If no digit may be repeated and each password must be at least 9 digits long, how many passwords are possible?

a. 9!+10!
b) 2!*10!
c) 9!*10!
d) 19!
e) 20!
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Jan 17, 2010 2:33 am

by Pilot » Fri Feb 18, 2011 12:52 pm
Hi, Night reader, thank you for your correction.

is my answer correct?

Do you mean, if the problem's condition were must be at least 8 digits we should have counted as 10P8+10P9+10P10?

Thanks!!!
Top
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Fri Feb 18, 2011 1:04 pm
it might be with simple combinatoric arrangement easily to omit several ways digits can be arranged, like AB and BA is one arrangement when we use combination and two arrangements with permutations. The effect of permutation use is counting (but not discounting) of possible arrangements.
Agree with 10P8*8!+10P9*9!+10P10*10! OR 3*10! :)
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Jan 17, 2010 2:33 am

by Pilot » Fri Feb 18, 2011 8:10 pm
Night reader, If you mean nPk=n!/(n-k)!, your answer is not correct.
Top
Post new topic Post Reply
• Page 1 of 1