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25 Dec 2011, 15:24
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Hi everyone,
I am trying to solve a hard question and i will be happy if I can use your help.
The question:
When creating a number from K digits with N cells.
What is the probability to that each of the digits will appear at least once?
for example, when k=3 (1, 2 and 3) and N cells the answer is:
[3^n-3-3(2^n-2)]/3^n
Thanks a lot! :-D
I am trying to solve a hard question and i will be happy if I can use your help.
The question:
When creating a number from K digits with N cells.
What is the probability to that each of the digits will appear at least once?
for example, when k=3 (1, 2 and 3) and N cells the answer is:
[3^n-3-3(2^n-2)]/3^n
Thanks a lot! :-D
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Thank you for understanding, and happy exploring!
P(each of the digits will appear at least once) + P(one digit will not appear at all) + P(two digits will not appear at all) + P(three digits will not appear at all) +.......+P(k-1 digits will not appear at all) = 1
=> P (each of the digits will appear at least once) = 1 - [P(one digit will not appear at all) + P(two digits will not appear at all) + P(three digits will not appear at all) +.......+P(k-1 digits will not appear at all)]
Let us refer to this relationship just derived as Rel 1.
Now, P(one digit will not appear at all) = (number of ways to choose one digit from k digits)*(number of ways to fill n cells using the remaining k-1 digits)/(number of ways to fill n cells using n digits)
=> P(one digit will not appear at all) = k*(k-1)^n/k^n
Similarly, P(two digits will not appear at all) = kC2 * (k-2)^n/k^n = [k(k-1)/2]*(k-2)^n/k^n
Therefore Rel 1 becomes
P(each of the digits will appear at least once) = 1 - [k*(k-1)^n/k^n + k(k-1)*(k-2)^n/2k^n +.....
Putting in k = 3 we get
P(each of the digits will appear at least once) = 1 -[(3*2^n)/3^n + 3/3^n] = (3^n -3*2^n - 3)/3^n
=[3^n + 3 -3(2^n + 2)]/3^n
Hope this helps.
=> P (each of the digits will appear at least once) = 1 - [P(one digit will not appear at all) + P(two digits will not appear at all) + P(three digits will not appear at all) +.......+P(k-1 digits will not appear at all)]
Let us refer to this relationship just derived as Rel 1.
Now, P(one digit will not appear at all) = (number of ways to choose one digit from k digits)*(number of ways to fill n cells using the remaining k-1 digits)/(number of ways to fill n cells using n digits)
=> P(one digit will not appear at all) = k*(k-1)^n/k^n
Similarly, P(two digits will not appear at all) = kC2 * (k-2)^n/k^n = [k(k-1)/2]*(k-2)^n/k^n
Therefore Rel 1 becomes
P(each of the digits will appear at least once) = 1 - [k*(k-1)^n/k^n + k(k-1)*(k-2)^n/2k^n +.....
Putting in k = 3 we get
P(each of the digits will appear at least once) = 1 -[(3*2^n)/3^n + 3/3^n] = (3^n -3*2^n - 3)/3^n
=[3^n + 3 -3(2^n + 2)]/3^n
Hope this helps.
26 Dec 2011, 14:31
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GyanOne
yes, it's helps a lot, thank you!
but because i need to create a general soulotion for any K that i want i need to transform it to a Series, But i don't know what the bolded C2 means, i understand that C2 = (k-1)/2 but what is the general formula for the C?
