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anonymous
 one year ago
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
 one year ago
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0permutation and combination unit. I know it's tricky.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Guys I know this is super difficult... Would really appreciate your insight

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3well, 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
 one year ago
Best ResponseYou've already chosen the best response.0Right. So that's a typical permutation without restrictions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer would be 36P36

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3well, 36P8, but you have the right idea

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah you are right.

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3ok, we can move on to the second question now "The password must feature both numbers and letters?"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would think that it has to do with 10 and 26?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need to take into account 36 options AND somehow play with 10 and 26 because password has to feature both

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3right, 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, so that's left would be the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me think about A.,,,

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3right, numbers only would be 10P8 and letters only would be 26P8 ready to keep going?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! So 36P8(26P8+10P8)=possible combinations including both letters and numbers.

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3haha, 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 361=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
 one year ago
Best ResponseYou've already chosen the best response.0What a concise explanation

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3well, it would actually be 26*[35P7]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Accounting all the possibilities for a letter right?

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3alright, 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
 one year ago
Best ResponseYou've already chosen the best response.0I think 26*(35P7)*10something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since I'd have to account for the numbers used until the last digit.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But knowing that is probably the trickiest.

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.326*(35P6)*10 would be my best guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would that number ever be repeated?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like say some number was already used before hand

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3yes, that's what I was thinking too

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3we have to account for the first and last character first, before we tackle the middle

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3so, we have 26 possibilities for the first character and 10 possibilities for the last character after those two characters are used up, we have 362 = 34 characters left for the 6 digits in the middle, giving us (26)*(34P6)*(10)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That 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
 one year ago
Best ResponseYou've already chosen the best response.0But within our limits of intuition yours seems to be the rightest.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n being how many numbers already used until then

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3right, I was thinking the same thing, it would be best to doublecheck with another, though @ganeshie8 @satellite76

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you summoned them?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you a math student in uni?

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3yeah, I'm only a sophomore though so I haven't done anything past Calculus/Linear Algebra though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nice! Is life full of discovery for you now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am stuck for a month or two in high school:(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi everyone Vocaloid and I had arrived at the answer 26*(34P6)*(10n) 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
 one year ago
Best ResponseYou've already chosen the best response.0I think your last answer is right Vocaloid:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No other way can it be expressed

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.3Right, 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
 one year ago
Best ResponseYou've already chosen the best response.0that's why I think it's n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think there is any degree of certainty until the last digit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because if all 6 digits be numbers, then the last one would be 26*(34PN)*(106)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So n be the determinant for the possiblities

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.026*(34P6)*(10) is right for last part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What if the numbers were used in until the last digit? That would limit how many numbers can be picked.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0good question, think of it this way : first pick first and last characters, then that ambiguity wont be there.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437017772324:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0After 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So happy now:) Thanks so much Ganeshie8!
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