cinar 3 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 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.

1. cinar

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

2. amistre64

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

3. amistre64

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

4. cinar
5. cinar

I thought that in that way..

6. amistre64

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

7. cinar

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

8. amistre64

yes

9. cinar

hmm, it makes sense now..

10. cinar

for c 52^7*10^3

11. 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

12. amistre64

im over thinking that one ...

13. cinar

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

14. 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

15. 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

16. 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?

17. amistre64

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

18. amistre64

i cant think thru that one :/

19. cinar

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

20. 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

21. cinar

typo 10^3

22. amistre64

10.9.8 = 720 ; /6 = 120 distinct ways

23. cinar

24. amistre64

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

25. cinar

yeah, brains do that sometimes (:

26. 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" ?

27. cinar

|dw:1346353576368:dw|

28. cinar

we have this situation for above case rigth

29. amistre64

something similar to that yes

30. cinar

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

31. cinar

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

32. cinar

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

33. cinar

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

34. 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 ....

35. amistre64

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

36. cinar

It seems logical..

37. cinar

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

38. amistre64

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

39. amistre64

or positions?

40. cinar

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

41. cinar

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

42. amistre64

sure, whats the other question :)

43. cinar

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..

44. amistre64

good luck :)

45. cinar

thanks, I need that (:

46. cinar

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)..

47. cinar

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..

48. cinar

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?

49. cinar

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

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