anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
a) I found like 36^10 but I'm not sure..
amistre64
  • amistre64
if alphanumeric defines just letters and numbers, then yes, there are 36 options for each position.
amistre64
  • amistre64
26*2 = 52 + 10 = 62, there are 62 options for each position with some modifications i believe for the b part

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
http://en.wikipedia.org/wiki/Alphanumeric
anonymous
  • anonymous
I thought that in that way..
amistre64
  • amistre64
62^10+62^11+62^12 is what im thinking for b such that abcdefghij differs from 0abcdefghij, 00abcdefghij
anonymous
  • anonymous
what does 10 to 12 alphanumeric characters mean, can we choose 10 alphanumeric, 11 alphanumeric or 12 alphanumeric ?
amistre64
  • amistre64
yes
anonymous
  • anonymous
hmm, it makes sense now..
anonymous
  • anonymous
for c 52^7*10^3
amistre64
  • 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
  • amistre64
im over thinking that one ...
anonymous
  • anonymous
I guess my solution is just valid for if first three passwords is number....right..
amistre64
  • 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
  • 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
  • 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
  • amistre64
3*10*10 = 300 ways .... and yes there are some double counts
amistre64
  • amistre64
i cant think thru that one :/
anonymous
  • anonymous
can we say for c 10^35*2^7*62^10
amistre64
  • 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
  • anonymous
typo 10^3
amistre64
  • amistre64
10.9.8 = 720 ; /6 = 120 distinct ways
anonymous
  • anonymous
but you added them previous solution, was your that solution wrong,,
amistre64
  • amistre64
the addition was a mistake on my part ... brain fell asleep while typing :)
anonymous
  • anonymous
yeah, brains do that sometimes (:
amistre64
  • 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
  • anonymous
|dw:1346353576368:dw|
anonymous
  • anonymous
we have this situation for above case rigth
amistre64
  • amistre64
something similar to that yes
anonymous
  • anonymous
I am gonna think over that later, you wanna see my other question, if you don't I don't mind (:
anonymous
  • anonymous
let assume we have small case for d) N={1,2,3,4} A={A,B,C,D,E,F}
anonymous
  • anonymous
and let say we have 2 number and password is 4 digit
anonymous
  • anonymous
if we can solve this we can also solve d (:
amistre64
  • 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
  • amistre64
i cant make out this instruction: "and let say we have 2 number and password is 4 digit"
anonymous
  • anonymous
It seems logical..
anonymous
  • anonymous
|dw:1346354587103:dw| I wanted ask actually this..
amistre64
  • amistre64
4*4*5*5, any restrictions on doubles?
amistre64
  • amistre64
or positions?
anonymous
  • anonymous
it is the same question with d just small elements, don't worry about it, your last solution works well.. thanks man..
anonymous
  • anonymous
my offer is still valid, you wanna see my other question (:
amistre64
  • amistre64
sure, whats the other question :)
anonymous
  • 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
  • amistre64
good luck :)
anonymous
  • anonymous
thanks, I need that (:
anonymous
  • 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
  • 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
  • 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
  • anonymous
720*52^7*C(10,3) there are 720 non-repeated numbers

Looking for something else?

Not the answer you are looking for? Search for more explanations.