BFMracing
General Category => General Board => Homework Haven => Topic started by: Racerrr on July 22, 2009, 12:45:43 AM
-
okay this isnt really homework but then again i hope this is the best section to ask if not then feel free 2 move it.. okay so i applied to an engineering school and the department gave me a test that i took. i still have my scratch people along with notes so i will reconstruct some of the questions asked. some of them are like obvious and stuff but some are not and some are tough that i had no idea. there were 50 questions so here are a few of them.
1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?
2) Suppose you are gambling with a friend of yours with a fair game, where the chance of winning and losing are 50/50. If you play 100 rounds, what's the probability you will win 50 times, no more & no less?
3) Write 2.3417417417417... as a fraction. For example, 0.9375 can be written as (30/32).
4) What is the solution to this differential equation: (2x-5y)dx + (-5x+3y^2)dy=0
5) Suppose a dealer for a casino rolls two standard dice. You saw one of the dice; it was a 2. What is the chance that the sum of the face value of the two dice adds up to 6?
6) Find a polynomial function f(x)=ax^2 + bx + c, where the graph passes though points (-1,14), (1,4), and (3,10).
7) Suppose your friend invites you to play a game. He shows you three cups, and exactly one of the cups has the prize. You don't know which one, and you have to tell him which cup you think conceals the prize. He will always reveal one of the empty cups which does not have the prize. So there are exactly two cups left: one with the prize, and one without the prize. He persuades you to switch cup. Should you switch, stay with your cup, or it doesn't really matter at all?
8 ) Suppose the profit a company makes can be model by the function: (X/10)*(1+Y/100)*(2+Z/80) subjected to the constraint that X+Y+Z=4000. What value of X,Y,Z will maximize the company's profit?
9) How many zeros are at the end of 200! (200 factorial)?
-
7) Suppose your friend invites you to play a game. He shows you three cups, and exactly one of the cups has the prize. You don't know which one, and you have to tell him which cup you think conceals the prize. He will always reveal one of the empty cups which does not have the prize. So there are exactly two cups left: one with the prize, and one without the prize. He persuades you to switch cup. Should you switch, stay with your cup, or it doesn't really matter at all?
Stay with the same one you originally chose. If you switch cups your probability goes down to 1/3. If you stay the same, your probability is still 2/3. (Or something like that).
I feel stupid now... I had it backwards ;D ;D
-
Give me some time and I'll try to answer as many of these as possible. Here's the answer to number 9:
9) How many zeros are at the end of 200! (200 factorial)?
Simply go through the numbers 1 - 200 and find all the ones that divide by ten. these are 10, 20, 30... 100 ... 190, 200. there are 22 tens in these numbers so there are 22 zeros at the end of 200!.
and to number 2:
2) Suppose you are gambling with a friend of yours with a fair game, where the chance of winning and losing are 50/50. If you play 100 rounds, what's the probability you will win 50 times, no more & no less?
Using the combination formula gives the answer as 50!*50!/100! which is probably too high for most calculators to give the exact answer, but is close to 9.9*10-30.
Also, navy's answer to number 7 is exactly wrong. You should always switch which gives probability 2/3 of winning (so why your friend's asking you to do it is beyond me). This is because you had a 2./3 chance of picking the wrong cup to start with. Google "The Monty Hall Problem" for full details.
-
Also, navy's answer to number 7 is exactly wrong. You should always switch which gives probability 2/3 of winning (so why your friend's asking you to do it is beyond me). This is because you had a 2./3 chance of picking the wrong cup to start with. Google "The Monty Hall Problem" for full details.
That's not what the movie 21 tells us :P
I could probably do 1 or 2 of these but I need time to work them out myself, and since I don't have time I won't try any just yet
-
1.
10(10)(26^9) This is wrong
2. Already done.
3. 23394/9990
4. -5xy + x2 + y3
5. 1/6
6. 2x2 - 5x + 7 (used matrix row reduction)
7. Already done.
8. This would take longer than the others, so I don't feel like doing it at 6:30 am.
9. You didn't quite do this correctly, 360. You forget that the 2 and 5 would make another zero, and there are other pairs like that as well. Would take longer as well. I'll just do it quickly even though it may be wrong by assuming each ending with 5 will have a corresponding multiple of 2, which would give 20 more I believe, giving 42 zeros.
Edit:
1. I'll just do this in summation notation to the best of my abilities.
Summation from n = 1 to n = 10 (n will essentially be number of numbers) of (10 C n)(10n)(2610 - n)
-
wow thank you for the quick reply
i suspected #7 was tricky so i guess the correct answer lol.
edson... how u compute the ans for #3 and #4?
and if any1 can work out #8 also, and is there a formula to solve it?
-
Number three is basically done by utilizing the repeating aspect of the .0417417417. For repeating stuff like that, it's just over something like 999 or 9 or 99, etc. So take 417/999 for .417417, then divide by 10 for .0417417417 and 417/9990. Then you just multiply 2.3 by 9990 and add it to 417.
Number four was a pretty simple differential equation, so I really just did it by inspection.
As for number 8, which I didn't do, I feel tempted to use Lagrangian Multipliers.
-
8. Yeah, having just taken Calc 3, that's a definite use of LaGrange Multipliers. Though I can't remember how you do that.
What school was this for, may I ask? I haven't heard of entrance exams like that around here.
-
Number three is basically done by utilizing the repeating aspect of the .0417417417. For repeating stuff like that, it's just over something like 999 or 9 or 99, etc. So take 417/999 for .417417, then divide by 10 for .0417417417 and 417/9990. Then you just multiply 2.3 by 9990 and add it to 417.
As for number 8, which I didn't do, I feel tempted to use Lagrangian Multipliers.
ahaa, i was like trying to figure out how to make something repeating... wow can't believe it was that simple. ty for the explanation. Can you calculate and tell me the answer for #8, also please?
Goalie, its some nonofficial benchmarking kind of thing to see where u stand and see if u need any leveling work. hopefully i did okay but i dont think that is the case but u never no.
-
This (http://en.wikipedia.org/wiki/Lagrangian_multipliers) should explain enough for you to try it out.
-
sry im dyslexic when it comes to reading somethin with bunch of math terminology, symbols, etc. cant figure out what that wiki thing is sayin. so how many steps does it take to do a problem like that and how much time for an avg. person? and how long u think it will take u to do it?
-
I'd have to review it really quickly, but I wouldn't think that it would take terribly long.
-
Whoops, Edison, I didn't pay enough attention there. Oo-er, all the 2*5 pairs... Also there are more fives buried in 25, 125, 200, etc., which means a bit more counting.
Well, 22 zeros so far from all the tens, then let's count all the fives (it's fairly safe to assume that we will have more than enough 2's for all these extra 5's I'd have thought since 64 and 128 give 13 2's alone). Exhaustive counting gives us 27 fives since 25, 75, 125, and 175 give 5 more fives over what you counted, while also we need to look at 150 and 50 which have an extra factor of 5 in each. So that allows for 27 more zeros or 49 in total.
-
I'm graduating with a Bachelor of Science in Mathematics this year, so hopefully I can help you with your questions.
1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?
1. It's a combinatorics (counting) problem. You basically calculate all the possible passwords that can be generated with 36 (26 letters + 10 digits) characters and then subtract the total number of passwords that consist of only letters. So the answer is 36^10-26^10. It's a lot easier than using summations and combinations to add up the sum of all possible cases like Edison did. Edison, you make something easy.... so complicated :LOL: I have my friends that are like that.
2. Going to disagree with the answer. It should be about 7.959%, not almost 0%. Each round, you have a 50-50 chance, so if you play a total of 100 rounds, then there are 2^100 unique possible outcomes, each equally likely with a probability of 1/(2^100). Since the person wins exactly 50 times, you have the multiply 1/(2^100) by the number of ways you can get exactly 50 wins out of 100 rounds, which is isomorphic to a combination problem. So the answer is 1/(2^100) * 100C50
Answer in LaTeX generated by mathurl.com:
(http://mathurl.com/kwwdbu.png)
3. Done by Edison
4. Done by Edison.
5. Going to disagree with Edison. It should be higher than 1/6. It's a Game Theory question where knowing a piece of information will dynamically change the probability. It also has basis in probability theory. I computed it to be 18.18%.
6. Done by Edison
7. Done by BFM_NavyJHk... backward.
8. @Racerrr.... there's not really a "formula" but it has only one simple constraint so it's easier to just set Fx=Fy=Fz=0 and solve for x,y,z instead of using Lagrange Multipliers. There are other easier methods if you ever take a course in Optimization Theory. If you still need a step-by-step guidance for this problem or any other, feel free to let me know.
9. I computed it to be 49 zeroes.... Number Theory certainly helps.
I'm still a n00b in math though, so if anyone thinks any of my answers are incorrect, then let me know.
-
On number 5, if you have two die, and one is a 2, there's only one possible scenario where you could get a total value of 6, which is with a 4 on the other die. The die are mutually exclusive so the first die has no impact on the second, making the chance 1/6, or the chance that it was a 4.
On the first one, I mainly just had misread it first off, thinking only one number, so I ended up just doing for two numbers and so on >.>
I knew something was missing in 360's for number two >.> It is 100 C 50 over 2100.
Figured there was a number theory way for 9.
-
On number 5, if you have two die, and one is a 2, there's only one possible scenario where you could get a total value of 6, which is with a 4 on the other die. The die are mutually exclusive so the first die has no impact on the second, making the chance 1/6, or the chance that it was a 4.
How confident are you in your answer, percentage-wise? :winkgrin:
-
Do you want me to make a program to stimulate it?
-
It sounds like you want to make a bet. If you are, I'll take it.
-
Simulated it just now 10 million times. And let's see how many times it added up to 6 given that the first one was 2... what do you know, 1666242, roughly 1/6 with a very low degree of uncertainty. :P
-
Simulated it just now 10 million times. And let's see how many times it added up to 6 given that the first one was 2... what do you know, 1666242, roughly 1/6 with a very low degree of uncertainty. :P
You might want to check your program for any logical error. Post your source code and I'll point it out for you.
Come on now Edison, all those advanced math classes you took in high school... you of all people should know that mathematics can be very counterintuitive and is full of paradoxes, so you can't always rely on "common sense" or intuition.. I'm sure you'll figure out why it can't be 1/6. If you give up, let me know and I'll explain why. :)
EDIT:
Here's a screen of my simulation I wrote in C++. As you can see, the probability gets closer and closer to my answer of 18.18% as the number of trials increase, in accordance with the laws of probability.
(http://i25.tinypic.com/122fbcp.png)
-
In my opinion, that's a different question than what I was answering, where it's the same dice every time that you see. They were probably aiming for your interpretation, but I still hate how they can never be specific as to what they want, like "What's the probability that one of the dice is a 2 and the sum is 6." Just ask exactly what you're looking for.
-
I believe the question is clearly asking... if someone rolls two standard dice, and you see one of the dice lands on a 2, then what is the probability the other dice landed on a 4.
Most people would say something like, "hmmmmm... the two dice are independent, and there's 6 sides on the other dice, and so there obviously must be a 1/6 chance of getting a 4 on the other dice." It would make perfect sense, except that it's not true. The question is do you know why? ;)
-
Except I read it as you are seeing the same dice every time in all these different situations, and not whichever has landed on two. So if you label each die as one and two, then it only counts as seeing die one as being a 2, but not seeing die two as a 2. I viewed it as in any scenario, it was a specific die you were seeing, not whichever one was a 2. That's just what the question meant to me. Thought I made it clear that I knew why you get 2/11 when I said what a clearer question would be with the whole fact that either die could be a 2.
-
You are complicating things again. If two dice are rolled together, if you see a 2, then the probability is very slighly greater the other dice is not a 2.
-
No it's not. You get 2/11 when you're looking at the chance that they add up to 6 with a 4 and a 2.
-
No it's not. You get 2/11 when you're looking at the chance that they add up to 6 with a 4 and a 2.
If you put 2/11 in your calculator, it will equal to 18.18%, my original answer from the beginning :)
-
I know that 2/11 is .181818181818. What I am saying is that one die rolling a 2 does not make the other die less likely to roll a 2. What I'm saying is that you're looking at the situation where either die can be a 2, while I'm looking at it as only a specific die can be a 2. So in your interpretation of either die possibly being a 2, it is 2/11, while in my interpretation of only a specific die, it is 1/6.
-
I know that 2/11 is .181818181818. What I am saying is that one die rolling a 2 does not make the other die less likely to roll a 2. What I'm saying is that you're looking at the situation where either die can be a 2, while I'm looking at it as only a specific die can be a 2. So in your interpretation of either die possibly being a 2, it is 2/11, while in my interpretation of only a specific die, it is 1/6.
Let's be realistic. If you go to a casino, and a dealer rolls two dice, you can't really tell if it was dice A or dice B that landed on a 2. But the fact that you saw a dice landed on a 2, you can predict the other dice is less likely to be a 2 also. The question isn't saying that the dealer rolls one dice at a time. It's rolled together, so they aren't independent of each other. There's no ambiguity here.
-
They are completely independent events. These dice aren't entangled. Your wording is basically saying that the probability for each number is 2/11 for every number except for 2, which is 1/11, which is not correct.
-
They are completely independent events. These dice aren't entangled. Your wording is basically saying that the probability for each number is 2/11 for every number except for 2, which is 1/11, which is not correct.
Actually, it is correct. Go to the store and buy 2 dice and conduct the experiment and you'll be amazed at the result. Since 16.67% and 18.18% are so close together, make sure your number of trials is sufficiently large enough so that statistically, you'll have a high level of confidence in your answer. Test your alternative hypothesis. :)
Or..... I guess taking some time off of Physics and spending more on Advanced Probability Theory wouldn't hurt.
-
I still don't think you see what I'm saying. In my interpretation, were I to test it using dice, I would have one red and one blue. I'd only look at the red one to see if it's a two. If it was, I would then see if the other is a four. In this way, I would get 1/6. In your interpretation, I would examine both dice, and if either was a two, then I'd look at the other and see if the other was a four. In this way, I would get 2/11. I ran simulations for both of these methods of looking for a two, and the first gave 1/6 while the second gave 2/11.
-
I do get exactly what you are saying. Your interpretation is wrong. The question says nothing about a red die and a blue die, looking a red die first, and then the blue die. The answer would be trivial if that was the case.
In your interpretation, I would examine both dice, and if either was a two, then I'd look at the other and see if the other was a four.
What would be the point of predicting the "other" die if someone already knows the value of both dice already? It makes no sense at all. Your wrong assumption of my interpretation will still produce the correct answer by coincidence though. However, the question is asking to predict the value of the "other" die by knowing the result of just one die that you see.
Oh, and if you ever gamble before, almost all of the time the dice will have the same color and are almost indistinguishable from one another.
-
My point is that in my mind, a theoretical simulation of this would have the dice rolling the same way exactly every time with different results, and only one specific die would be seen. It's as if you were to repeat the same exact motions of the dice every time, but they result in different numbers. In no way is it an incorrect interpretation of the problem, even if it may not be the interpretation that the askers meant. And the part you quoted was me simplifying what the difference between our two interpretations even more, since it didn't seem like you understood what I was saying, and I don't know that you do now for sure. I have a completely correct interpretation of your interpretation, which is that either dice could have been seen, and that you can't restrict yourself to the possibility of only ever seeing one of the dice. It's not by coincidence that they have the same value, as the meaning behind them is the same; they have the exact same method of simulation in a computer program. And so what if that would be trivial? Some of the other questions presented were, in my mind, trivial as well.
My presentation of different colored dice was to illustrate my interpretation more clearly, but you're taking it somewhat too literally. To simulate it by hand, since you can't have a way to make sure that they person happens to see a specific die every time, they'd have to be different colors for my interpretation, as trivial as it may be. The triviality of a question has never stopped people from asking it.
-
I wasn't taking what you said literally; I was merely trying to point out a flaw in your logic. That's all :winkgrin:
Anyways...
Original Question:
5) Suppose a dealer for a casino rolls two standard dice. You saw one of the dice; it was a 2. What is the chance that the sum of the face value of the two dice adds up to 6?
I'd prefer to calculate things in my head, but as they always say, "When in doubt, list them all out." Of course, this problem has a small sample space so listing them is reasonable.
So the question says that two dice are rolled together and "you saw one of the dice; it was a 2." Since there 2 dice, and there 6 possible values on each die, then there are 36 possible outcomes. I don't think anyone will disagree on that, for it's just basic math. So here are sample space:
{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}
Since you saw a die with a "2", then one of these 11 outcomes must have occured:
{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}
And from those 11 possible outcomes, only 2 of them add up to 6.
{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}
That is why your answer of 1/6 is incorrect. The problem is very counterintuitive, just like the Monty Hall problem. But I have to give it to you, you have good debate skills.
-
Except you still don't see what I'm trying to say. In my interpretation, where only a specific die can be seen as having a two, this is what you get:
{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}
There are 6 scenarios where the first die is a 2, and only one of them where the second die is a 4, giving 1/6. The scenarios where the second die is a 2 do not count in my interpretation as they are not on the die that you saw.
I already knew how you got your answer. The point that I've been trying to get across is that we're answering different questions.
-
I figured that to resolve this once and for all I'll just go back to basics.
What is the probability that two dice add up to 6 given that one of them is a two?
Then using that lovely Bayes' formula for these circumstances gives:
P(6 given a 2) = P(6)*P(a 2 given 6)/P(a 2) = (5/36)*(2/5)/(11/36) = 2/11.
QED.
-
I've never denied that. I explicitly stated earlier that the probability that two dice add up to 6 given that one of them is a 2 is 2/11. I can find the quote if you want. What I've said multiple times and will continue saying is that I read (present tense) the problem as asking if a specific die is 2, and in any simulation, it is that die that is 2. In my interpretation, it's not "one of them is a two," but, "this specific one is a two, and the other can be a two or not, but I don't really care." I'll say it one more time, we are answering two separate questions here.
-
Edison, you have taken so many positions throughout this thread that you seem to confuse people. You seem to be all over the place. You frist claimed the answer was 1/6, then when I said it was 2/11, you asked me if I wanted to see you "make a program to simulate it" to show my answer of 2/11 is incorrect, then you claimed to have the "completely correct interpretation of [my] interpretation" that I would have to "examine both die" to calculate the answer, which I showed to not be true.
EDIT: I missed you post, three360. Thank you for giving a new perspective. <3
-
That's not what I said at all. I was saying that checking both dice for a two is the same problem in essence as the probability that you roll a 2 and a 4. I never said you had to, but that it is the exact same question in essence. The only time I've changed a position is when I realized that you interpreted the question itself differently, resulting in a different, non-trivial answer. I'm done, explaining this bores me >.>
Btw, the way 360 did it was the way I solved it for your interpretation as well.
-
It's okay, Edison. One day you'll be able to correct me when I'm wrong.
Oh and the PM you sent me.... LOL I'm up for it anytime :P
-
No one's wrong here >.>
-
One way of explaining why I think the 2/11 interpretation is right is that, in the context of the other questions on the paper, it is clear that a non-trivial answer is required. I initially just looked at the question and said "1/6" but that requires no thought at all. This is a paper with some hard questions on and I can't honestly think that they would expect you to be able to answer a question by just looking at it - or, at least, that there would be a question on dice asking essentially, "what is the probability of getting a 4 on a fair die?", a question I'd hope that most 11-year-olds could answer. By introducing the 2nd die then it is clearly supposed to complicate the maths involved or lengthen the working.
-
2, 3, and 4 seem pretty trivial to me, as does 6 if you know how to fit polynomials to specific points.
-
Guys, seriously.
Both of you are right. You just looked at the question in different ways. If you rolled two random dice, the probability that you rolled a 4 after rolling a 2 is 1/6. The probability that they equal 6 GIVEN THAT EITHER is a 2 is 2/11. You're both right in different ways. Move on, seriously. LOL! :LOL:
I think 2/11 is the corret answer on the paper though. :interesting:
-
Guys, seriously.
Both of you are right. You just looked at the question in different ways. If you rolled two random dice, the probability that you rolled a 4 after rolling a 2 is 1/6. The probability that they equal 6 GIVEN THAT EITHER is a 2 is 2/11. You're both right in different ways. Move on, seriously. LOL! :LOL:
I think 2/11 is the corret answer on the paper though. :interesting:
That's what I've been saying in almost every single post D:
-
2, 3, and 4 seem pretty trivial to me, as does 6 if you know how to fit polynomials to specific points.
I think three360 was talking about triviality based on the perception of the average Joe Schmoe IMO. Why is the rest not trivial to you?
I haven't done #6 yet, but can the solution be obtained by solving this 3x4 matrix below using the Jordan-Gaussian algorithm to put it in reduced row echelon form?
1 -1 1 14
1 1 1 4
9 3 1 10
-
Everything is trivial when you know how to do it.
For #6 I think you start by setting up 3 simultaneous equations then doing with them how you please. I imagine the matrix form suggested is neatest but I'd probably pick the Ax=b form and look at it like:
1 -1 1 a = 14
1 1 1 b 4
9 3 1 c 10
And try Gaussian elimination or something like it. SO I'm solving a 3*3 matrix which I believe is slightly easier.
-
2, 3, and 4 seem pretty trivial to me, as does 6 if you know how to fit polynomials to specific points.
I think three360 was talking about triviality based on the perception of the average Joe Schmoe IMO. Why is the rest not trivial to you?
I was going with the ones that I think Joe Schmoe should know and consider trivial the most, of course the key word is should but >.>
-
Use the reasonable person test!
-
Personally, I think it's easier to just RREF a a 3x4 matrix than solving for Ax=b
I RREF'ed
1 -1 1 14
1 1 1 4
9 3 1 10
and got
1 0 0 2
0 1 0 -5
0 0 1 7
I think it's right.
-
I refereed it and checked the points for my original answer, which I think is the same as yours.
-
I refereed it and checked the points for my original answer, which I think is the same as yours.
Yup, it appears so.
-
what is a differential equation???????!!!!!!!! :doh:
-
A differential equation is pretty much exactly what it says it is - an equation invloving differentials. So I expect you're really asking what a differential is. I'll try to answer that with a practical example rather than mathematically.
Suppose you have a hot cup of tea and you're interested in how long it will take to cool down. For that information you need to know how hot it is, how hot the surrounding environment is and how fast heat moves from one to the other. But it's not unreasonable to assume that the hotter the tea is, the faster heat moves from it to the room. Hence not so much the time taken to cool is important but the rate at which the tea cools down. This depends on the difference between the tea's temperature, T, and the room's temperature, R, and importantly it is not a constant because hot things cool down more rapidly that colder things. Since at all times the quantity (T-R) is changing, the rate of cooling is given by d(T-R)/dt where d/dt is a differential and expresses the fact that conditions are always in a state of change. It's then possible to solve the resulting equation (time taken to cool is proportiional to d(T-R)/dt ) using the rules of calculus.
Differentials come from the problem of trying to solve problems that involve curves when it's not possible to use straight lines, rectangles and so on. Then the first step was to approximate to a straight line and then find better and better approximations, when the exact answer comes from considering a sraight line with no length. Kind of confusing I know, but that's sort of what happens although it's a little more complicated than that.
-
1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?
ok thats a math c question
26+10 is 36 possible characters taking into account the number
35x34x33x32x31x30x29x28x27
-
I don't think that's right. 35x34x33... assumes you only use each letter/number once. But you could have aaaaabbbb9 where the a's and b's repeat.
I think it's more like
10 (all the numbers) * 36 (letters/numbers) to the ninth power = 10*36*36*36*36*36*36*36*36*36
or does the answer involve factorials...hmmm...
-
1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?
I believe that it would be 10x36x36x36x36x36x36x36x36x36
The x's mean multiply (I never liked *).
Eeach number there (there are 10) represents the sample space (don't quote me on correct use of terms and definitions).
The password must contain at least one number, meaning there are 10 to choose from (0,1,2,3,4,5,6,7,8,9).
Then the rest can be either a lower case letter or a number, meaning there are 36 to choose from (q,w,e,r,t,y,u,i,o,p,a,s,d,f,g,h,j,k,l,z,x,c,v,b,n,m,1,2,3,4,5,6,7,8,9,0).
Therefore, the answer is 1015599566684160, or one quadrillion, fifteen trillion, five hundred and ninety-nine billion, five hundred and sixty-six million, six hundred and eighty-four thousand, one hundred and sixty
-
I don't think that's right. 35x34x33... assumes you only use each letter/number once. But you could have aaaaabbbb9 where the a's and b's repeat.
I think it's more like
10 (all the numbers) * 36 (letters/numbers) to the ninth power = 10*36*36*36*36*36*36*36*36*36
or does the answer involve factorials...hmmm...
ur right
i was thinking non replacement
-
This topic is old, and I am far too lazy to go back and look for what answers were posted. The best way to go about this is to find all without the restriction on at least one number and then take away all those that don't have a number. That is, you have 36^10-26^10 = (36^5-26^5)(36^5+26^5) = 48,584,800x72,347,552 = 3,514,991,344,409,600. The factoring was an attempt to not get errors due to tenth powers being too large, but I needed Mathematica anyways. The problem with just trying 10x35^9 is that you restrict where you are placing the number. This would seem to multiply your value by 10 again for each of the places it can be placed, but then you're counting some possibilities more than once. For example, 12b1 changed to 2b11 gives you something you already would have counted. Thus in cases where you have "at least", you want to start with total cases and subtract those away that don't satisfy the "at least" factor.
-
This topic is old, and I am far too lazy to go back and look for what answers were posted. The best way to go about this is to find all without the restriction on at least one number and then take away all those that don't have a number. That is, you have 36^10-26^10 = (36^5-26^5)(36^5+26^5) = 48,584,800x72,347,552 = 3,514,991,344,409,600. The factoring was an attempt to not get errors due to tenth powers being too large, but I needed Mathematica anyways. The problem with just trying 10x35^9 is that you restrict where you are placing the number. This would seem to multiply your value by 10 again for each of the places it can be placed, but then you're counting some possibilities more than once. For example, 12b1 changed to 2b11 gives you something you already would have counted. Thus in cases where you have "at least", you want to start with total cases and subtract those away that don't satisfy the "at least" factor.
I was waiting for you to say that.
-
Everything is trivial when you know how to do it.
I love that, best thing I've heard all day aha