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 2 years ago
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 casesensitive.
(b) The password has 10 to 12 alphanumeric characters, and is casesensitive.
(c) The password has 10 casesensitive alphanumeric characters, and
exactly 3 of them are numerical.
(d) The password has 10 casesensitive alphanumeric characters, exactly 3 of
them are numerical, and no numerical character is repeated.
 2 years ago
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 casesensitive. (b) The password has 10 to 12 alphanumeric characters, and is casesensitive. (c) The password has 10 casesensitive alphanumeric characters, and exactly 3 of them are numerical. (d) The password has 10 casesensitive alphanumeric characters, exactly 3 of them are numerical, and no numerical character is repeated.

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cinar
 2 years ago
Best ResponseYou've already chosen the best response.0a) I found like 36^10 but I'm not sure..

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1if alphanumeric defines just letters and numbers, then yes, there are 36 options for each position.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.126*2 = 52 + 10 = 62, there are 62 options for each position with some modifications i believe for the b part

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0I thought that in that way..

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.162^10+62^11+62^12 is what im thinking for b such that abcdefghij differs from 0abcdefghij, 00abcdefghij

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0what does 10 to 12 alphanumeric characters mean, can we choose 10 alphanumeric, 11 alphanumeric or 12 alphanumeric ?

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0hmm, it makes sense now..

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1for 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
 2 years ago
Best ResponseYou've already chosen the best response.1im over thinking that one ...

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0I guess my solution is just valid for if first three passwords is number....right..

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1yes, 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
 2 years ago
Best ResponseYou've already chosen the best response.1the 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
 2 years ago
Best ResponseYou've already chosen the best response.1001 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
 2 years ago
Best ResponseYou've already chosen the best response.13*10*10 = 300 ways .... and yes there are some double counts

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1i cant think thru that one :/

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0can we say for c 10^35*2^7*62^10

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1for c; 10^3*52^7 is valid for one positional set up there are \[\frac{10!}{3!7!}\]distinct ways to set it up

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.110.9.8 = 720 ; /6 = 120 distinct ways

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0but you added them previous solution, was your that solution wrong,,

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1the addition was a mistake on my part ... brain fell asleep while typing :)

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, brains do that sometimes (:

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1001 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" ?

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0we have this situation for above case rigth

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1something similar to that yes

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0I am gonna think over that later, you wanna see my other question, if you don't I don't mind (:

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0let assume we have small case for d) N={1,2,3,4} A={A,B,C,D,E,F}

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0and let say we have 2 number and password is 4 digit

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0if we can solve this we can also solve d (:

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1000 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
 2 years ago
Best ResponseYou've already chosen the best response.1i cant make out this instruction: "and let say we have 2 number and password is 4 digit"

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1346354587103:dw I wanted ask actually this..

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.14*4*5*5, any restrictions on doubles?

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0it is the same question with d just small elements, don't worry about it, your last solution works well.. thanks man..

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0my offer is still valid, you wanna see my other question (:

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1sure, whats the other question :)

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0I 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..

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0I 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)..

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0for d) we are looking for number of nonrepeated 3 digit number, right? to do that, 10*10*10= repeated is allowed, 10*9*8=repeated is not allowed so, 10*10*1010*9*8= 280 =number of nonrepeated 3 digit number..

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0so 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?

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0720*52^7*C(10,3) there are 720 nonrepeated numbers
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