Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

I thought that in that way..

Answer 6

62^10+62^11+62^12 is what im thinking for b such that

abcdefghij differs from 0abcdefghij, 00abcdefghij

Answer 7

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

Answer 8

yes

Answer 9

hmm, it makes sense now..

Answer 10

for c 52^7*10^3

Answer 11

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

Answer 12

im over thinking that one ...

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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?

Answer 17

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

Answer 18

i cant think thru that one :/

Answer 19

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

Answer 20

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

Answer 21

typo 10^3

Answer 22

10.9.8 = 720 ; /6 = 120 distinct ways

Answer 23

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

Answer 24

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

Answer 25

yeah, brains do that sometimes (:

Answer 26

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" ?