password codes
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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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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?
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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I think answer would be 10C4 * 4! = 5040.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.)
What is OA?
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well! If I follow this answer then it can be break down to (9*8*7)/10*10*10 = 0.504.nh8404052006 wrote:OA : [spoiler]10P4 /10^4[/spoiler]
Isn't there something wrong with the answer stated above by you?
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Selecting 4 alphabets from 10 alphabets = 10C4.kishore wrote:I think answer would be 10C4 * 4! = 5040.???
Can you explain how did u come to this conclusion?
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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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.
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.