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a) I found like 36^10 but I'm not sure..

if alphanumeric defines just letters and numbers, then yes, there are 36 options for each position.

http://en.wikipedia.org/wiki/Alphanumeric

I thought that in that way..

yes

hmm, it makes sense now..

for c 52^7*10^3

im over thinking that one ...

I guess my solution is just valid for if first three passwords is number....right..

3*10*10 = 300 ways .... and yes there are some double counts

i cant think thru that one :/

can we say for c 10^35*2^7*62^10

typo 10^3

10.9.8 = 720 ; /6 = 120 distinct ways

but you added them previous solution, was your that solution wrong,,

the addition was a mistake on my part ... brain fell asleep while typing :)

yeah, brains do that sometimes (:

|dw:1346353576368:dw|

we have this situation for above case rigth

something similar to that yes

I am gonna think over that later, you wanna see my other question, if you don't I don't mind (:

let assume we have small case for d) N={1,2,3,4} A={A,B,C,D,E,F}

and let say we have 2 number and password is 4 digit

if we can solve this we can also solve d (:

i cant make out this instruction: "and let say we have 2 number and password is 4 digit"

It seems logical..

|dw:1346354587103:dw|
I wanted ask actually this..

4*4*5*5, any restrictions on doubles?

or positions?

my offer is still valid, you wanna see my other question (:

sure, whats the other question :)

good luck :)

thanks, I need that (:

so answer is for d)
10^3*52^7*C(10,3)-280*52^7*C(10,3)
can we say this? or is this true?

720*52^7*C(10,3)
there are 720 non-repeated numbers