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password codes

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by nh8404052006 » Sun Dec 28, 2008 6:44 pm
A software program is designed to generate password codes using alphabet from A to J, how many 4-alphabet password codes can the program generate? (please note: the alphabets are not allowed to repeat itself.)
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password codes

by kishore » Sun Dec 28, 2008 7:50 pm
A software program is designed to generate password codes using alphabet from A to J, how many 4-alphabet password codes can the program generate? (please note: the alphabets are not allowed to repeat itself.)?


Answer: A to J there are 10 aplhabets.

Each password is 4 digits length.

Every time you create a password you should pick 4 different letters so that you avoid repetition.

10C4 = 10!/6! * 4! = 210

210 is the correct answer

May I know the IMO?
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Re: password codes

by Sachindh » Sun Dec 28, 2008 7:56 pm
nh8404052006 wrote:A software program is designed to generate password codes using alphabet from A to J, how many 4-alphabet password codes can the program generate? (please note: the alphabets are not allowed to repeat itself.)
I think answer would be 10C4 * 4! = 5040.

What is OA?
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by nh8404052006 » Sun Dec 28, 2008 8:02 pm
OA : [spoiler]10P4 /10^4[/spoiler]
I am not sure the OA is correct or not.
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by Sachindh » Sun Dec 28, 2008 8:12 pm
nh8404052006 wrote:OA : [spoiler]10P4 /10^4[/spoiler]
well! If I follow this answer then it can be break down to (9*8*7)/10*10*10 = 0.504.

Isn't there something wrong with the answer stated above by you?
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by kishore » Sun Dec 28, 2008 8:35 pm
I think answer would be 10C4 * 4! = 5040.???


Can you explain how did u come to this conclusion?
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by Sachindh » Sun Dec 28, 2008 8:38 pm
kishore wrote:I think answer would be 10C4 * 4! = 5040.???


Can you explain how did u come to this conclusion?
Selecting 4 alphabets from 10 alphabets = 10C4.
Arranging selected 4 alphabets to created different passcodes = 4!
Total no of ways = 10C4 * 4!

Let me know if there is some thing wrong in the approach.
Thanks
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by vittalgmat » Sun Dec 28, 2008 11:43 pm
This is a permutation problem.
I am so happy to see so many ways that this problem can be solved.
I solved it analytically.

There are 10 alphabets and 4 positions.
mark them as
1----2----3----4

1st position can be filled by any 10 alphabets. So 10 ways
2nd position can be filled by any 9 alphabets. So 9 ways
3d " " "" 8 alphabets . So 8 ways
4th position needs 7 ways.

So total ways is a product of all these = 5040 ways.
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by cramya » Mon Dec 29, 2008 12:08 am
Agree 100% with some posters on 5040

10*9*8*7
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