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