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01-10-2011
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#1 |
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Ma ghayro
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Math dudes, ta3o ta nshouf
Okay, i really suck at math and i really need to solve this.
I have all these characters and i want to know how many up to 3,4,5,6 combinations can i have for the following 62 characters (Repetition is allowed). 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMN OPQRSTUVWXYZ 4 characters eg : could be the following: D f9 vz1 m6H2 Remember repetition is allowed eg: D3fD
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01-10-2011
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#2 |
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62 to the power 4 = 14776336
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01-10-2011
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#3 |
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Beyond the Code
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what he said, here why:
entry: _ _ _ _ 1st digit can be any of the 62 characters 2nd digit the same, same for the 3rd and the 4th so its 62x62x62x62 no repetition it would be: 62 choices for the 1st digit x 61 choices for the 2nd digit x 60 choices for the 3rd digit x 59 choices for the 4th digit. each time the choices has been decreased by 1 because it's taken by the preceding digit.
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01-10-2011
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#4 |
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What do you mean by up to 3,4,5,6?
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01-10-2011
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#5 |
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he is saying that for for n permutations it is actually:
for n permutations (here 4) what he wants is perm(4) + perm (3) + perm(2) not only perm(4)
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01-11-2011
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#6 |
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the answer is
62^1+62^2+...+62^N where N=3,4,5,6 (or whatever you choose it)
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01-11-2011
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#7 |
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the answer is:
2  Thank You ...
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| The Following User Says Thank You to abousoun For This Useful Post: | TAREK® (01-11-2011) |
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01-12-2011
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combination of 3: 62^3
comb of 4: 62^4 of 5 : 62^5 of 6: 62^6 |
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| dudes, math, nshouf, ta3o |
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