I need some help these discrete math questions:
For each part below, ﬁnd the number of passwords that ﬁt the condition.
(a) The password has 10 alphanumeric characters, but is not case-sensitive.
(b) The password has 10 to 12 alphanumeric characters, and is case-sensitive.
(c) The password has 10 case-sensitive alphanumeric characters, and
exactly 3 of them are numerical.
(d) The password has 10 case-sensitive alphanumeric characters, exactly 3 of
them are numerical, and no numerical character is repeated.

- anonymous

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- anonymous

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

- amistre64

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

- amistre64

26*2 = 52 + 10 = 62, there are 62 options for each position with some modifications i believe for the b part

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## More answers

- anonymous

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

- anonymous

I thought that in that way..

- amistre64

62^10+62^11+62^12 is what im thinking for b such that
abcdefghij differs from 0abcdefghij, 00abcdefghij

- anonymous

what does 10 to 12 alphanumeric characters mean, can we choose 10 alphanumeric, 11 alphanumeric or 12 alphanumeric ?

- amistre64

yes

- anonymous

hmm, it makes sense now..

- anonymous

for c 52^7*10^3

- amistre64

for c im thinking:
N,N,N,a,a,a,a,a,a,a given that N has 10 and a has 52
\[10C3~(10^3+52^7)\]but that wouldnt account for 333 being the same as 333 would it

- amistre64

im over thinking that one ...

- anonymous

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

- amistre64

yes, since there is not a position defined, we would have to take that value and multiply it by the number of ways the positions can change

- amistre64

the last one is the same as the one before it, we just have to subtract the set of "same numbers"
NNN
NNn
NnN
nNN
there appears to be 4 ways that that set can be arranged, so all we would have to do is figure out how many are in the set to begin with

- amistre64

001 991
010 ... 919 ; 10 ways
100 199
002 992
020 ... 929
200 299
...
009 999
090 ... 999
900 999
-----
10 ways
is it safe to say there are 100 ways to get duplicated numbers?
or am i introducing a double count?

- amistre64

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

- amistre64

i cant think thru that one :/

- anonymous

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

- amistre64

for c; 10^3*52^7 is valid for one positional set up
there are \[\frac{10!}{3!7!}\]distinct ways to set it up

- anonymous

typo 10^3

- amistre64

10.9.8 = 720 ; /6 = 120 distinct ways

- anonymous

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

- amistre64

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

- anonymous

yeah, brains do that sometimes (:

- amistre64

001 991
010 ... 919 ; 10 ways ... but 111,111,111 suggests that we need to adjust
100 199 by subtracting 2
3*10 -2 = 28, there are 10 ways to run thru this making: c - 280
can we see any issues with this idea for "d" ?

- anonymous

|dw:1346353576368:dw|

- anonymous

we have this situation for above case rigth

- amistre64

something similar to that yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

000 001 002 003 004 005 006 007 008 009
010 020 030 040 050 060 070 080 090
100 200 300 400 500 600 700 800 900
110 111 112 113 114 115 116 117 118 119
101 121 131 141 151 161 171 181 191
011 211 311 411 511 611 711 811 911
220 221 222 223 224 225 226 227 228 229
202 212 232 242 252 262 272 282 292
022 122 322 422 522 622 722 822 922
330 331 332 333 334 335 336 337 338 339
303 313 323 343 353 363 373 383 393
033 133 233 433 533 633 733 833 933
i cant see an issue with it ....

- amistre64

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

- anonymous

It seems logical..

- anonymous

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

- amistre64

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

- amistre64

or positions?

- anonymous

it is the same question with d just small elements, don't worry about it, your last solution works well.. thanks man..

- anonymous

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

- amistre64

sure, whats the other question :)

- anonymous

I gotta go now, I'll try to solve myself at home tonight, and if I have some trouble I'll post it here and I'll call your name, thank you so much again, see you later..

- amistre64

good luck :)

- anonymous

thanks, I need that (:

- anonymous

I got now question c, C(10,3) for position of this 3 numbers in 10 slot, but we can also use C(10,7) for position of the letters in 10 slot right. It odes not matter we are choosing number or letter, and also we know this C(10,3)=C(10,7)..

- anonymous

for d) we are looking for number of non-repeated 3 digit number, right?
to do that, 10*10*10= repeated is allowed, 10*9*8=repeated is not allowed
so, 10*10*10-10*9*8= 280 =number of non-repeated 3 digit number..

- anonymous

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?

- anonymous

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

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