Question

A password length can be 6 to 8 characters long, where each character is an upper case letter or a digit. each password must contain at least one digit. how many possible passwords are there?

Answer 1

k so to make this easier... lets look at just a 6 character long password

the book says you should find the total number of possibilities and subtract away all of the possible passwords that are only alpha
which would be 36^6 - 26^6

why couldn't you just force 1 of the slots to be numeric and do 36^5 * 10???

Answer 2

Help?

Answer 3

your version would be correct only if it were 5 alphanumeric characters followed by a digit at the end.

Answer 4

you've lost the degree of freedom of where the digit may appear, e.g. that you are allowed to have a character as the last if a number appears earlier.

Answer 5

help me please? you are the only one that seems to be on and i would very much be thankful for someone to help me please.

Answer 6

wait why though? you are using the product rule and where the slot is doesn't matter

Answer 7

?

Answer 8

doing 36^4 * 10 * 36 is the same thing

Answer 9

i mean with my problem above?

Answer 10

your rule gets all possible combinations of letters and numbers but undercounts the ways they can be permuted

Answer 11

36^5 * 10 is only correct if there is exactly one position that is required to have a digit, and all the other positions can be either letters or numbers

Answer 12

but isn't that what it's asking?
A password length can be 6 to 8 characters long, where each character is an upper case letter or a digit. each password must contain at least one digit.

at least 1 would mean 1 position required to be a digit and the rest could be anything, no?

Answer 13

bleh, easy stuff just don't want to go into the test and make a mistake like this

Answer 14

for a thought experiment, do this problem: how many valid passwords are there if positions 1 through 5 are all required to be a letter or number, and position 6 can only be a number?

Answer 15

604661760 no? isn't it the same thing?

Answer 16

36^5 * 10, right? that's the same answer you wanted to give for the first problem. But the password "1AAAAA" is valid for the first problem, and invalid for the second problem. thus the number of valid passwords can not be the same.

Answer 17

ahhh ok wait so i haven't read the chapter on combinatorics yet. Remember doing it a while back, but it's been a long time.

Answer 18

ok so the thought experiment is a permutation, in which order does matter right?

Answer 19

and the problem from the book is a combination?

Answer 20

36^5 * 10 is the number of valid passwords if there is one particular position (e.g. position #6) that has the digit-only restriction. this undercounts the number of valid passwords in the original problem, where the digit can be in *any* position.

Answer 21

i thought there might be a way to "patch" your formula by multiplying it by 5, to account for the number of places it's legal to put the digit, but that's broken too - it overcounts. i think the only way to do it correctly is the way the book said.

Answer 22

ah ok i think i kind of get it now. think it's just going to take some reading into the chapter on combinations/permutations to really get a handle on the subtleties to these problems

Answer 23

thank you very much for your help. really appreciate it

Answer 24

no problem

Answer 25

i found another way to get it

Answer 26

36^5*10 + 36^4*10*26 + 36^3*10*26^2 + 36^2*10*26^3 + 36*10*26^4 +
10*26^5

Answer 27

you will never get it with just multiplication

Answer 28

you want me to explain it?

Answer 29

what the... i dont understand why you have to do all of that

Answer 30

that gives the right answer but i'm also perplexed as to why

Answer 31

and why would you do *26 + *26^2... and so on vs doing *10 + *10^2... etc

Answer 32

say the first number comes in the first slot. the next 5 can be anything. that's 10*36^5

Answer 33

k

Answer 34

ohhh i see, then let's say hypothetically the first is not a number, but the second is, and the rest can be anything

Answer 35

if the first number comes in the 2nd slot, the first character has to be a letter. and the last 4 can be anything. 26*10*36^4

Answer 36

ahhhhhhhhhh

Answer 37

or the first two are both not numbers, and the third is, and the rest can be anyhthing

Answer 38

if the first number comes in the 3rd slot, the 1st 2 characters have to be letters, and the last 3 can be anything 26^2*10*36^3

Answer 39

haha that's clever, if unwieldy

Answer 40

yes, it's easier to do it the way the book did

Answer 41

yeah i agree but your way makes a lot more sense why the method works

Answer 42

ah very nice very nice. well thanks a ton fellas