Question

HELP PLEASE FAN, MEDAL AND TESTIMONIAL
Idi is creating a password for a website that has some strict requirements. The password must be 8 characters. Numbers and letters may be used, but may not be repeated.
How many diffferent passwords are possible?
How many are possible if the following restrictions are enforced;
The password must feature both numbers and letters?
The password must start with a letter?
The password must start with a letter and end with a number?

Answer 1

Do you know?

Answer 2

permutation and combination unit. I know it's tricky.

Answer 3

Guys I know this is super difficult... Would really appreciate your insight

Answer 4

well, let's start off with the first question

"how many different passwords are possible?"

we can choose any of the 10 digits or any of the 26 letters (the question doesn't specify whether the password is case sensitive or not, so I'm going to go ahead and use 26). this gives us a possible of 10+26 = 36 possibilities for each of the 8 characters in the password

so, for the first character we have 36 choices
for the second character, we have 35 choices since we already used one
for the third character, we have 34 choices, and so on, until we get to the last character

total number of possibilities = (36)*(35)*(34)*(33)*(32)*(31)*(30)*(29)

Answer 5

Right. So that's a typical permutation without restrictions.

Answer 6

So the answer would be 36P36

Answer 7

well, 36P8, but you have the right idea

Answer 8

Oh yeah you are right.

Answer 9

ok, we can move on to the second question now

"The password must feature both numbers and letters?"

Answer 10

I would think that it has to do with 10 and 26?

Answer 11

I need to take into account 36 options AND somehow play with 10 and 26 because password has to feature both

Answer 12

right, we want to find out

A. how many passwords only use numbers
and
B. how many passwords only use letters

then, we want to figure out

36P8 - A - B

Answer 13

Oh, so that's left would be the answer.

Answer 14

right!

Answer 15

Let me think about A.,,,

Answer 16

26P8?

Answer 17

10P*?

Answer 18

10P8

Answer 19

right, numbers only would be 10P8 and letters only would be 26P8

ready to keep going?

Answer 20

Yes! So 36P8-(26P8+10P8)=possible combinations including both letters and numbers.

Answer 21

Wow you are so good

Answer 22

haha, thanks ~

next question is "The password must start with a letter"

so, for the first character, we have 26 possibilities since we can only use letters
for the second character, we have 36-1=35 possibilities, since we already used one of the letters
then 34, then 33, and so on until we get to the 8th character

Answer 23

35P8 IT IS!!!!

Answer 24

What a concise explanation

Answer 25

:):)

Answer 26

well, it would actually be 26*[35P7]

Answer 27

Accounting all the possibilities for a letter right?

Answer 28

Yes I get it.

Answer 29

alright, last one is the trickiest, it might take me a minute to get the answer

"The password must start with a letter and end with a number?"

Answer 30

I think 26*(35P7)*10something

Answer 31

Since I'd have to account for the numbers used until the last digit.

Answer 32

But knowing that is probably the trickiest.

Answer 33

26*(35P6)*10 would be my best guess

Answer 34

however...

Answer 35

Would that number ever be repeated?

Answer 36

Like say some number was already used before hand

Answer 37

yes, that's what I was thinking too

Answer 38

we have to account for the first and last character first, before we tackle the middle

Answer 39

ok

Answer 40

-n?

Answer 41

so, we have 26 possibilities for the first character and 10 possibilities for the last character

after those two characters are used up, we have 36-2 = 34 characters left for the 6 digits in the middle, giving us

(26)*(34P6)*(10)

Answer 42

That looks about the closest to being right? I don't quite get how we are supposed to know if numbers may have been used in the (34P6) part in which case the last digit option would decrease instead of full 10?

Answer 43

But within our limits of intuition yours seems to be the rightest.

Answer 44

(26)*(34P6)(10-N)?

Answer 45

n being how many numbers already used until then

Answer 46

right, I was thinking the same thing, it would be best to double-check with another, though

@ganeshie8 @satellite76

Answer 47

so you summoned them?

Answer 48

You are so nice:)

Answer 49

Are you a math student in uni?

Answer 50

yeah, I'm only a sophomore though so I haven't done anything past Calculus/Linear Algebra though

Answer 51

Nice! Is life full of discovery for you now?

Answer 52

I am stuck for a month or two in high school:(

Answer 53

Hi everyone Vocaloid and I had arrived at the answer 26*(34P6)*(10-n) where n is the number of numbers used in making 8 digit password composed of letters and numbers. I would really appreciate your insights as well

Answer 54

I think your last answer is right Vocaloid:)

Answer 55

No other way can it be expressed

Answer 56

Right, that's my best guess. I'm still a bit uncertain about the whole "how do we account for situations where numbers have already been used" but eh

Answer 57

that's why I think it's n

Answer 58

I don't think there is any degree of certainty until the last digit

Answer 59

because if all 6 digits be numbers, then the last one would be 26*(34PN)*(10-6)

Answer 60

26*(34P6)*(10-6)

Answer 61

So n be the determinant for the possiblities

Answer 62

26*(34P6)*(10)

is right for last part

Answer 63

not n?

Answer 64

nope

Answer 65

What if the numbers were used in until the last digit? That would limit how many numbers can be picked.

Answer 66

good question, think of it this way :

first pick first and last characters, then that ambiguity wont be there.

Answer 67

Oh

Answer 68

you are right .

Answer 69

In the start, we have 26+10 = 36 characters to choose from. But we want the first character to be a letter, so there are 26 ways to choose first character :