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?

- anonymous

- chestercat

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- anonymous

Do you know?

- anonymous

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

- anonymous

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

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## More answers

- Vocaloid

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)

- anonymous

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

- anonymous

So the answer would be 36P36

- Vocaloid

well, 36P8, but you have the right idea

- anonymous

Oh yeah you are right.

- Vocaloid

ok, we can move on to the second question now
"The password must feature both numbers and letters?"

- anonymous

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

- anonymous

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

- Vocaloid

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

- anonymous

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

- Vocaloid

right!

- anonymous

Let me think about A.,,,

- anonymous

26P8?

- anonymous

10P*?

- anonymous

10P8

- Vocaloid

right, numbers only would be 10P8 and letters only would be 26P8
ready to keep going?

- anonymous

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

- anonymous

Wow you are so good

- Vocaloid

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

- anonymous

35P8 IT IS!!!!

- anonymous

What a concise explanation

- anonymous

:):)

- Vocaloid

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

- anonymous

Accounting all the possibilities for a letter right?

- anonymous

Yes I get it.

- Vocaloid

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?"

- anonymous

I think 26*(35P7)*10something

- anonymous

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

- anonymous

But knowing that is probably the trickiest.

- Vocaloid

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

- Vocaloid

however...

- anonymous

Would that number ever be repeated?

- anonymous

Like say some number was already used before hand

- Vocaloid

yes, that's what I was thinking too

- Vocaloid

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

- anonymous

ok

- anonymous

-n?

- Vocaloid

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)

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

n being how many numbers already used until then

- Vocaloid

right, I was thinking the same thing, it would be best to double-check with another, though
@ganeshie8 @satellite76

- anonymous

so you summoned them?

- anonymous

You are so nice:)

- anonymous

Are you a math student in uni?

- Vocaloid

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

- anonymous

Nice! Is life full of discovery for you now?

- anonymous

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

- anonymous

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

- anonymous

I think your last answer is right Vocaloid:)

- anonymous

No other way can it be expressed

- Vocaloid

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

- anonymous

that's why I think it's n

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

So n be the determinant for the possiblities

- ganeshie8

26*(34P6)*(10)
is right for last part

- anonymous

not n?

- ganeshie8

nope

- anonymous

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

- ganeshie8

good question, think of it this way :
first pick first and last characters, then that ambiguity wont be there.

- anonymous

Oh

- anonymous

you are right .

- ganeshie8

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 :
|dw:1437017702350:dw|

- anonymous

Yes

- anonymous

|dw:1437017772324:dw|

- ganeshie8

After that, we choose the last character instead of the second.
We have 35 characters to pick from, but we want the last character to be a number. so we have 10 ways to choose a number for last character :
|dw:1437017797071:dw|

- ganeshie8

I see you get it :)

- anonymous

So happy now:) Thanks so much Ganeshie8!

- anonymous

Thanks everyone!

- ganeshie8

np, thank Vocal

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