anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Do you know?
anonymous
  • anonymous
permutation and combination unit. I know it's tricky.
anonymous
  • anonymous
Guys I know this is super difficult... Would really appreciate your insight

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Vocaloid
  • 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
  • anonymous
Right. So that's a typical permutation without restrictions.
anonymous
  • anonymous
So the answer would be 36P36
Vocaloid
  • Vocaloid
well, 36P8, but you have the right idea
anonymous
  • anonymous
Oh yeah you are right.
Vocaloid
  • Vocaloid
ok, we can move on to the second question now "The password must feature both numbers and letters?"
anonymous
  • anonymous
I would think that it has to do with 10 and 26?
anonymous
  • anonymous
I need to take into account 36 options AND somehow play with 10 and 26 because password has to feature both
Vocaloid
  • 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
  • anonymous
Oh, so that's left would be the answer.
Vocaloid
  • Vocaloid
right!
anonymous
  • anonymous
Let me think about A.,,,
anonymous
  • anonymous
26P8?
anonymous
  • anonymous
10P*?
anonymous
  • anonymous
10P8
Vocaloid
  • Vocaloid
right, numbers only would be 10P8 and letters only would be 26P8 ready to keep going?
anonymous
  • anonymous
Yes! So 36P8-(26P8+10P8)=possible combinations including both letters and numbers.
anonymous
  • anonymous
Wow you are so good
Vocaloid
  • 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
  • anonymous
35P8 IT IS!!!!
anonymous
  • anonymous
What a concise explanation
anonymous
  • anonymous
:):)
Vocaloid
  • Vocaloid
well, it would actually be 26*[35P7]
anonymous
  • anonymous
Accounting all the possibilities for a letter right?
anonymous
  • anonymous
Yes I get it.
Vocaloid
  • 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
  • anonymous
I think 26*(35P7)*10something
anonymous
  • anonymous
Since I'd have to account for the numbers used until the last digit.
anonymous
  • anonymous
But knowing that is probably the trickiest.
Vocaloid
  • Vocaloid
26*(35P6)*10 would be my best guess
Vocaloid
  • Vocaloid
however...
anonymous
  • anonymous
Would that number ever be repeated?
anonymous
  • anonymous
Like say some number was already used before hand
Vocaloid
  • Vocaloid
yes, that's what I was thinking too
Vocaloid
  • Vocaloid
we have to account for the first and last character first, before we tackle the middle
anonymous
  • anonymous
ok
anonymous
  • anonymous
-n?
Vocaloid
  • 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
  • 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
  • anonymous
But within our limits of intuition yours seems to be the rightest.
anonymous
  • anonymous
(26)*(34P6)(10-N)?
anonymous
  • anonymous
n being how many numbers already used until then
Vocaloid
  • Vocaloid
right, I was thinking the same thing, it would be best to double-check with another, though @ganeshie8 @satellite76
anonymous
  • anonymous
so you summoned them?
anonymous
  • anonymous
You are so nice:)
anonymous
  • anonymous
Are you a math student in uni?
Vocaloid
  • Vocaloid
yeah, I'm only a sophomore though so I haven't done anything past Calculus/Linear Algebra though
anonymous
  • anonymous
Nice! Is life full of discovery for you now?
anonymous
  • anonymous
I am stuck for a month or two in high school:(
anonymous
  • 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
  • anonymous
I think your last answer is right Vocaloid:)
anonymous
  • anonymous
No other way can it be expressed
Vocaloid
  • 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
  • anonymous
that's why I think it's n
anonymous
  • anonymous
I don't think there is any degree of certainty until the last digit
anonymous
  • anonymous
because if all 6 digits be numbers, then the last one would be 26*(34PN)*(10-6)
anonymous
  • anonymous
26*(34P6)*(10-6)
anonymous
  • anonymous
So n be the determinant for the possiblities
ganeshie8
  • ganeshie8
26*(34P6)*(10) is right for last part
anonymous
  • anonymous
not n?
ganeshie8
  • ganeshie8
nope
anonymous
  • anonymous
What if the numbers were used in until the last digit? That would limit how many numbers can be picked.
ganeshie8
  • ganeshie8
good question, think of it this way : first pick first and last characters, then that ambiguity wont be there.
anonymous
  • anonymous
Oh
anonymous
  • anonymous
you are right .
ganeshie8
  • 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
  • anonymous
Yes
anonymous
  • anonymous
|dw:1437017772324:dw|
ganeshie8
  • 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
  • ganeshie8
I see you get it :)
anonymous
  • anonymous
So happy now:) Thanks so much Ganeshie8!
anonymous
  • anonymous
Thanks everyone!
ganeshie8
  • ganeshie8
np, thank Vocal

Looking for something else?

Not the answer you are looking for? Search for more explanations.