Sir I have a doubt , if we consider there are 10 possible ways to fill the first place in that situation 0 will also be included so we would get 3 digit number not 4 digit number
Wow thanks, you made this so easy to understand. Was wrapping my head around trying to figure out how many possible 7 digit codes could I generate from a pool of 27 digits using repetition.
a person wants to have a four digit password with the first digit greater than 5.how many different paswords the person can create using the digits 0 to 9
If we allow repetition: there are 4 choices for digit 1, then 10 choices for each of the remaining 3 digits of the password. If we forbid repetition: there are 4 choices for digit 1, then 9 choices for digit 2, etc.
I have been having a hard time understanding this topic 'til I watched this video. Thank you Sir. You just earned a follower. Hoping you could make a video about the same topic but with different restriction like only divisible by a number.
Probably a lot of truth to that! Thanks for watching and let me know if you ever have any questions! While I'm not wearing my Santa hat in videos these days, I am going into the woods for a lot of them!
Very glad it helped! Thanks for watching and let me know if you ever have any video requests! I’m gonna try to keep existing and making math lessons as long as I can :)
3:49 can you explain how can there be 10 possibilities of putting first digit , because if we put 0 as first digit, it will make whole a 3 digit no. (Eg. 0329. Zero neglected, which make it a 3 digit rather than 4) So I think there should be 9 ways of putting first digit rather than 10
Is there a formula I can use in Excel to provide me a list of individual 4-digit combinations without repetition? I've forgotten the combination to a Masterlock key box that uses pushbuttons that can only be used once. Thanks in advance!
Sir, we are asked to make a four digit number using 0 to 9 i.e 10 options . So, if we make these with or without repetition the number 0 if used as a thousands digit number would not make a four digit number. So, sir I think there are 9 possible options form the thousands place.
The digits 3 to 9 are available to set a 4 digit code for you locker at school. a) How many different codes are possible? b) Calculate the probability if the digits may not be repeated
Thanks for watching and for the question! For part a you just have to think...how many digits to you have to choose? And how many options do you have for each digit? Then multiply. For part b, I am not sure what is being asked. Calculate the probability of what?
From your homework problem "Thus, there are 5*4*3*2 = 120 4-digit codes using the even digits and not allowing repetition.", can you show this in your P(x,y) format as well?
Interesting effects going on there 😊 At around 5:11 in the video, you mentioned the formula P(10,4)... Now is it possible to have a situation in Permutations where the formula becomes P(4,10) ? I'm sorry if this comes across as a stupid question. I'm trying to figure out Permutations for a project. Thanks in advance.
Thanks for watching and for the question, it's a very good question! For these counting problems, we would never have use for P(4, 10), or in general for P(n, r) where r is greater than n, because r objects cannot be selected from n objects if r is greater than n. The only way that could be done is if repetition was allowed, in which case a simpler formula works. For example, if we were selecting 4 objects from a collection of 10, but repetition was allowed, then the number of ways we could do that would be 10^4. For example, imagine we were selecting 4 flavors of ice cream to go on a cone, from a total of 10 flavors. Hope that helps!
I have a request.... could you take a look at this to advise how this would be worked? I am a little confused. You have been asked to create license plates for the City of Asheville, NC. Suppose license? plates are to have three letters followed by three digits, and the first digit cannot be zero. -How many different license plates can be made (replacement is okay)? -How many different license plates can be made if letters cannot be duplicated? Thank you in advance
Sir, at 7:01 you have taken total of 6 possibilities for the thousandth's place, however those possibilities also include 0, which would effectively make the proposed 4-digit number in a 3-digit number. What do we do?
Thanks for watching and for the question! It entirely depends on the context of the problem. That's why this video has the word "codes" in the title. You might say 0123 isn't really a four digit number, but it certainly is a four digit code, as you might use for a lock combination. Does that help? If you wanted to disallow zero you'd have to specify that in the problem.
There are only 9 possibilities for thousands place for a four digit number because if 0 is placed at thousands place then it will not be a four digit number anymore.
What about a four digit code with 0-9, but working out how many of the same number? e.g. four matching digits there's 10 possible codes, so how many have three matching numbers in any position? Or two digits etc
I play Killer Sudoku -- I want to know which numbers 1-9, given n codes, when added together add-up to say 14 -- so, four empty cells that add up to 14. Since the 4 empty cells are in the same cage they cannot repeat. How do we solve that?
In the case of a lock combination the code 1, 9, 2, 4 is different from 9, 1, 4, 2 but in the case of a lottery win those two combinations are the same. What would be the formula for the lottery case?
Thanks for watching and the question! It depends on exactly what you're asking. But the most important first question would be how many different lottery combinations are there? If it consists of four numbers then we need to count the number of ways we can choose four numbers from the ten digits, disregarding order and without replacement (meaning once we select 9, we cannot select it again). Here is a video on the math necessary for this: th-cam.com/video/wKgiSUnbLxE/w-d-xo.html
Can anyone help me with a harder question - For the first two digits use an odd number between 39 and 100, for the last two digits use a multiple of 10, repetition is allowed work out the total number of codes which can be made using the method
Kev Cook I play Killer Sudoku -- I want to know which numbers 1-9, given n codes, when added together add-up to say 14 -- so, four empty cells that add up to 14. Since the 4 empty cells are in the same cage they cannot repeat. How do we solve that?
I'd like a formula to spit out all sum combinations....I don't want the number of permutations, I want the possible combinations. Like -- 4 boxes that add up to 15 (no repeat)....9+3+2+1, would be one, then 8+4+2+1, would be another.....
Thanks for watching and for the question! We can solve the problem with similar thinking to what we used for the lesson. How many digits could go in the first spot of our ID number? We could use 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9, so there are 10 total options. If every digit in the code is the same, then how many options do we have for the second, third, and fourth spots? Answer those questions, multiply the number of possibilities we have for each spot together, and that is your answer! Let me know if you're still not sure!
Thanks for watching and for your question! Let me make sure I understand. We're considering all positive integers that have exactly 9 distinct digits, like 123456789, 304678912, 541290873, those sorts of integers? And we want to count how many there are, is that the question?
Wait sir.. let me just comment here the whole question, and by the way thank you for replying on my comment such a big help for me to deal with my subject..
Consider all positive integers with 9 different digits. (note that 0 must not be on the first digit.) a.) How many are even? d.) How many are greater than 890,700,000? b.) How many are odd? e.) How many are less than 567,400,000? c.) How may are divisible by 5? f.) How many have the digits 1, 2 & 3?
That's the whole question sir.. no specific given right? That's the thing that I can't understand because there's no specific given besides the positive integers which is 9
Sir I have a question!! If 2,3,5,7,8 are the numbers from which the four digit code is formed..if 5 is the first digit of the code and the next two digits of the code are repeated numbers..what are the possible codes that can be formed???
i was wondering about this subject, turns out i was right. apparently googld wasn't right this time(i searched up 5 digit code with 0-9 and it gave me a wrong answer). So thank you
Thanks for watching and good question! It may help to think of this backwards, from the fourth digit to the first digit. How many choices do you have for the fourth digit? Then for the third? And so on.
Thanks for watching and I might be able to help clarify if you give me a timestamp (when in the video are you asking about?) and let me know where you're thinking the 24 comes from?
How do you write something like this into an expression? I understand where the 10 ^P 4 comes from and know how to calculate that with the function and stat feature on my calculator, but I am needing to learn how to write equations such as this into an equation and cannot grasp the concept. Hopeful. Jenn
I would imagine it's the same concept. You have 30 possible numbers to choose from and 4 places to put those numbers. 30*30*30*30=810000 I think the rules apply across the board no matter the variations that you are given, you just have to substitute how many possible combinations there are within every digits place.
Thanks for watching, I am not sure I understand your question. If we are talking about numerical passcode, the number of digits is the password length. Or are you asking about if we have, for example, a 20 digit passcode but only have 10 digits? Then, of course, if we disallow repetition, there are 0 possible passcodes.
I have a 4 digit code I’m trying to figure out. I know the last two digits _ _ 6 9. Numbers 0-9 and there’s no repetition. How can I figure it @Wrath of Math
Thanks for watching Valeria and the question you asked has many solutions. Are there perhaps some details you've left out? If we know it is a 4-digit code, the last two digits are 69, and no repetition is allowed, then 1269, 1369, 4369, are all examples of valid codes, and there are many more. Are we trying to count the number of options? Is that the question?
Thanks for watching and I'm sorry to hear of your physical discomfort! if I can be of any help to clear anything up and relieve the pain please let me know! Keep up the hard work!
Haha, I don't drink so it was pure water and CO2, I promise! I like seeing my decorated math set in this video! Kinda makes me wanna decorate my set again for the season...maybe something spooky 😉 Thanks for watching!
If we use 0 in ist digit place it will become 3 digit number ( because of insignificant) so there should be 9 possibilities in ist digit place i think?🤔
Thanks for watching, I'll copy a reply I made on this video to a similar comment. It entirely depends on the context of the problem. That's why this video has the word "codes" in the title. You might say 0123 isn't really a four digit number, but it certainly is a four digit code, as you might use for a lock combination. Does that help? If you wanted to disallow zero you'd have to specify that in the problem.
My problem is a bit different. I've got letters I, R, N and P and I've got to figure out all the combinations but without repetitions so I-I-I-R and R-I-I-I counts as one as they are the technically the same. I've computed 34 combinations but I don't know if I've missed any. I get a bit lost trying to use a brute force algorithm and I need to ensure I've got all the combinations. I had to do this for a 2 code length and 3 code length also. The 2 code length had 10 combinations and the 3 code one had 19 so I feel like I might have missed one or more there also. Anyone that could help, it would be greatly appreciated. Thanks.
Thanks for watching Alison! Could you clarify your question? You say "without repetition" but then list I-I-I-R as an option, which repeats I three times. If you're making combinations of four letters from I R N and P, without repetition, then there is only one option, which is to include all four of the letters. I imagine I am misunderstanding your question.
@@WrathofMath It's okay, I think 34 combinations is right. Bascially there can be repetition of the value but not the overall combination. So like your in video, a code of 9900 is fine but a code of 0099 would class as repetition as they contain the same numbers just in a different order. Using your method, there was 256 combinations but many of those will have repetition within them. So 9900 is fine but then I can't include 9009 or 0099 or 0990 because they each contain the same contents. I don't know if that helps? So for the two code length, I had [I-I, I-R, I-N, R-R, R-N, I-S, N-N, R-S, S-N and S-S] I can't also have [R-I, N-I, N-R, S-I, S-R and N-S] because they are the same as previous values, albeit in a different order. Hope that makes more sense? Tried a different method for brute force and got the same 34 combinations so I presume that is right but wondered if there was a maths trick to finding this out for certain, in case I have to increase to 5, 6, 7 and other code lengths in future. Thanks
Thanks for watching Nicole, and that sounds a like a fun restriction to mix up the problem. I'll try to do a video on that soon. Think about how forcing the number to be less than 7000 restricts our choices of digits!
This video isn't published yet, but with this link you can watch it early! th-cam.com/video/eoy8fHiWomU/w-d-xo.html EDIT: HD version of the video hasn't processed yet, so if you watch it shortly after I post this comment its resolution will be a bit low
Thanks for watching and that's a great question! If we were strictly making numbers by selecting digits, then that would be a very important detail for us to point out! We would only have 9 choices for the first digit, because if - for example - we are trying to make 4-digit numbers, if we start with 0 it is really a 3 digit number. In this video though we weren't talking about numbers specifically, we were just talking about 4-digit codes, like you might use for a password or PIN - where place value doesn't really exist. Does that clear it up?
Although logically it is correct to use 0 as our first number, it isn't really the case. Most of the time 0 is not counted as the start of a 2 (or more) digit number so therefore in the case of the 4-digit number 0 can't be used on the first slot so that means there are only 9 out of 10 possibilities.
Thanks for watching and I would agree if I had framed the problem as "4-digit numbers", but I specifically framed it as "4-digit codes" because codes can start with anything, the sort you might set as your phone's passcode. But your point is an important one to keep in mind, sometimes we will be counting numbers - not codes - and then we do have to exclude 0 from the first digit's options!
If you declare with your mouth, Jesus is Lord and believe in your heart that God raised him from the dead, you will be saved Romans10:9,Repent and Accept Jesus Christ as your Lord and Savior before its too late,or there will be consequences
Thanks for watching, Siddha, and good question! I answer your question in this new video: th-cam.com/video/lP4kTCDB58E/w-d-xo.html around the 44:38 mark
Take a break from studying to listen to some fire math raps: th-cam.com/video/2icLUHNioDk/w-d-xo.html
Sir I have a doubt , if we consider there are 10 possible ways to fill the first place in that situation 0 will also be included so we would get 3 digit number not 4 digit number
Wow thanks, you made this so easy to understand. Was wrapping my head around trying to figure out how many possible 7 digit codes could I generate from a pool of 27 digits using repetition.
So glad it helped! You're welcome and thanks for watching!
27^7
Thank you so much for showing both examples of repeat digits and no repeats. :)
No problem, thanks for watching!
Thank you for making it easy to understand
You're very welcome and thanks for watching!
a person wants to have a four digit password with the first digit greater than 5.how many different paswords the person can create using the digits 0 to 9
If we allow repetition: there are 4 choices for digit 1, then 10 choices for each of the remaining 3 digits of the password.
If we forbid repetition: there are 4 choices for digit 1, then 9 choices for digit 2, etc.
I have been having a hard time understanding this topic 'til I watched this video. Thank you Sir. You just earned a follower. Hoping you could make a video about the same topic but with different restriction like only divisible by a number.
Where is sir from? 👀
Online Class Graduation Speeches Be Like:
First of all I wanna thank TH-cam....
Probably a lot of truth to that! Thanks for watching and let me know if you ever have any questions! While I'm not wearing my Santa hat in videos these days, I am going into the woods for a lot of them!
Thank god you exist. Super duper helpful
Very glad it helped! Thanks for watching and let me know if you ever have any video requests! I’m gonna try to keep existing and making math lessons as long as I can :)
3:49 can you explain how can there be 10 possibilities of putting first digit , because if we put 0 as first digit, it will make whole a 3 digit no. (Eg. 0329. Zero neglected, which make it a 3 digit rather than 4)
So I think there should be 9 ways of putting first digit rather than 10
Not if you use a combination lock with a "zero" as first number.🐦
You made it easy.. thank you sir
Glad it helped, you're welcome and thanks for watching!
Is there a formula I can use in Excel to provide me a list of individual 4-digit combinations without repetition? I've forgotten the combination to a Masterlock key box that uses pushbuttons that can only be used once. Thanks in advance!
Sir, we are asked to make a four digit number using 0 to 9 i.e 10 options . So, if we make these with or without repetition the number 0 if used as a thousands digit number would not make a four digit number. So, sir I think there are 9 possible options form the thousands place.
There are still 10000 because the combinations are from 0000 to 9999
This video really helped me thank you!
You just solved what I was looking for
Glad to help, thanks for watching!
but what if i'm trying to find the probability that number has at least one digit repeated?
thank you for makin it easy to undersanding
My pleasure, thanks for watching!
The digits 3 to 9 are available to set a 4 digit code for you locker at school.
a) How many different codes are possible?
b) Calculate the probability if the digits may not be repeated
Thanks for watching and for the question! For part a you just have to think...how many digits to you have to choose? And how many options do you have for each digit? Then multiply.
For part b, I am not sure what is being asked. Calculate the probability of what?
From your homework problem "Thus, there are 5*4*3*2 = 120 4-digit codes using the even digits and not allowing repetition.", can you show this in your P(x,y) format as well?
Good day please assist on this one......1. how many code words over a,b,c,d of length 20 contains exactly 10 a's?
Interesting effects going on there 😊
At around 5:11 in the video, you mentioned the formula P(10,4)... Now is it possible to have a situation in Permutations where the formula becomes P(4,10) ?
I'm sorry if this comes across as a stupid question. I'm trying to figure out Permutations for a project. Thanks in advance.
Thanks for watching and for the question, it's a very good question! For these counting problems, we would never have use for P(4, 10), or in general for P(n, r) where r is greater than n, because r objects cannot be selected from n objects if r is greater than n. The only way that could be done is if repetition was allowed, in which case a simpler formula works. For example, if we were selecting 4 objects from a collection of 10, but repetition was allowed, then the number of ways we could do that would be 10^4. For example, imagine we were selecting 4 flavors of ice cream to go on a cone, from a total of 10 flavors. Hope that helps!
I have a request.... could you take a look at this to advise how this would be worked? I am a little confused.
You have been asked to create license plates for the City of Asheville, NC. Suppose license?
plates are to have three letters followed by three digits, and the first digit cannot be zero.
-How many different license plates can be made (replacement is okay)?
-How many different license plates can be made if letters cannot be duplicated?
Thank you in advance
Thank you best video!!
You're welcome, thanks so much for watching, D'Anna!!
I really struggle with permutation. Thank you this really helped me
So glad it helped, thanks for watching!
Thank you ✨
Sir, at 7:01 you have taken total of 6 possibilities for the thousandth's place, however those possibilities also include 0, which would effectively make the proposed 4-digit number in a 3-digit number.
What do we do?
Thanks for watching and for the question! It entirely depends on the context of the problem. That's why this video has the word "codes" in the title. You might say 0123 isn't really a four digit number, but it certainly is a four digit code, as you might use for a lock combination. Does that help? If you wanted to disallow zero you'd have to specify that in the problem.
There are only 9 possibilities for thousands place for a four digit number because if 0 is placed at thousands place then it will not be a four digit number anymore.
Correct!
What about a four digit code with 0-9, but working out how many of the same number? e.g. four matching digits there's 10 possible codes, so how many have three matching numbers in any position? Or two digits etc
Thank youu so muchhhh
My pleasure, thanks for watching!
Thank you very much
My pleasure, thanks for watching and let me know if you ever have any video requests!
In thousands place 9p1=9
In hundreds place 10p1=10
In tens place 10p1=10
In unit place 10p1=10
Therefore
Total =9×10×10×10=9000
I play Killer Sudoku -- I want to know which numbers 1-9, given n codes, when added together add-up to say 14 -- so, four empty cells that add up to 14. Since the 4 empty cells are in the same cage they cannot repeat. How do we solve that?
In the case of a lock combination the code 1, 9, 2, 4 is different from 9, 1, 4, 2 but in the case of a lottery win those two combinations are the same. What would be the formula for the lottery case?
Thanks for watching and the question! It depends on exactly what you're asking. But the most important first question would be how many different lottery combinations are there? If it consists of four numbers then we need to count the number of ways we can choose four numbers from the ten digits, disregarding order and without replacement (meaning once we select 9, we cannot select it again). Here is a video on the math necessary for this: th-cam.com/video/wKgiSUnbLxE/w-d-xo.html
@@WrathofMath
Thanks for that. Yes, that is the formula I needed. The divide by K! was the step I was looking for.
Thanks for the info ☺️🙂
You're welcome, thanks for watching!
Can anyone help me with a harder question - For the first two digits use an odd number between 39 and 100, for the last two digits use a multiple of 10, repetition is allowed work out the total number of codes which can be made using the method
Bro doubt , if we need a 4 digit number, if we count zero in first place then it will become 3 digit number. Can we count zero
In a group of twenty (20)boys, sixteen (16)play hockey and two (2)do not play either game.How many play both game?
Kev Cook I play Killer Sudoku -- I want to know which numbers 1-9, given n codes, when added together add-up to say 14 -- so, four empty cells that add up to 14. Since the 4 empty cells are in the same cage they cannot repeat. How do we solve that?
I'd like a formula to spit out all sum combinations....I don't want the number of permutations, I want the possible combinations. Like -- 4 boxes that add up to 15 (no repeat)....9+3+2+1, would be one, then 8+4+2+1, would be another.....
How many possible ID number are there in which all 4 digits are the same?
Thanks for watching and for the question! We can solve the problem with similar thinking to what we used for the lesson. How many digits could go in the first spot of our ID number? We could use 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9, so there are 10 total options. If every digit in the code is the same, then how many options do we have for the second, third, and fourth spots?
Answer those questions, multiply the number of possibilities we have for each spot together, and that is your answer! Let me know if you're still not sure!
Hi I have a question hope you can help me..Consider all positive integers with 9 different digits. (note that 0 must not be on the
first digit.)
Thanks for watching and for your question! Let me make sure I understand. We're considering all positive integers that have exactly 9 distinct digits, like 123456789, 304678912, 541290873, those sorts of integers? And we want to count how many there are, is that the question?
Wait sir.. let me just comment here the whole question, and by the way thank you for replying on my comment such a big help for me to deal with my subject..
Consider all positive integers with 9 different digits. (note that 0 must not be on the
first digit.)
a.) How many are even? d.) How many are greater than 890,700,000?
b.) How many are odd? e.) How many are less than 567,400,000?
c.) How may are divisible by 5? f.) How many have the digits 1, 2 & 3?
That's the whole question sir.. no specific given right? That's the thing that I can't understand because there's no specific given besides the positive integers which is 9
Thank you
My pleasure, thanks for watching!
How many odd numbers can be in a 4digit number, or how many combinations can be made by 3 odd and 1 even number (4digits)
I get it i do my module but i don't understand how thanks for your example 💖💖💖
Sir I have a question!!
If 2,3,5,7,8 are the numbers from which the four digit code is formed..if 5 is the first digit of the code and the next two digits of the code are repeated numbers..what are the possible codes that can be formed???
What is the least 4 digit with no repeating numbers that is divisible by 5????
i was wondering about this subject, turns out i was right. apparently googld wasn't right this time(i searched up 5 digit code with 0-9 and it gave me a wrong answer). So thank you
How many 4 digit even numbers can be formed from the digits 1, 3, 5, 6, 8 and 9 if repetition of digits is not allowed?
Thanks for watching and good question! It may help to think of this backwards, from the fourth digit to the first digit. How many choices do you have for the fourth digit? Then for the third? And so on.
I’m trying to crack my radio code. It’s a4 digit code using the numbers 1,2,3,4,5 and repetition is ok ?
i understand what you're saying but shouldn't you multiply all of that by 24 for the with repetition one?
Thanks for watching and I might be able to help clarify if you give me a timestamp (when in the video are you asking about?) and let me know where you're thinking the 24 comes from?
How do you write something like this into an expression? I understand where the 10 ^P 4 comes from and know how to calculate that with the function and stat feature on my calculator, but I am needing to learn how to write equations such as this into an equation and cannot grasp the concept. Hopeful. Jenn
0 to 9 four digit combinations with out repetition .how to get all the 9*9*8*7 numbers
How many non repeating 4 digit codes can be set up using 0 thru 9 and A thru Z?
Meaning I do not want the same digit twice in a single code
What happens when you chang the options like your trying to find the the amount of possibility's of 1-30 in 4 number combinations
I would imagine it's the same concept. You have 30 possible numbers to choose from and 4 places to put those numbers.
30*30*30*30=810000
I think the rules apply across the board no matter the variations that you are given, you just have to substitute how many possible combinations there are within every digits place.
What if the amount of digits is less than the password length?
Thanks for watching, I am not sure I understand your question. If we are talking about numerical passcode, the number of digits is the password length. Or are you asking about if we have, for example, a 20 digit passcode but only have 10 digits? Then, of course, if we disallow repetition, there are 0 possible passcodes.
@@WrathofMath thanks for responding! yes that’s my question. But what if you allow repetition of numbers, but want unique permutations???
MY PROFESSOR SAID ANS IS 9000 IF 0 - 9 OF 4 DIGIT COMBINATION IF WE AVOID 0 ON FISRT POSSITION HOW PLEASE GIVE SOLUTION....
Hello sir can you use excel using your 4 digit logic that would be very easy to understand for student like me
Thanks
Thanks!
My pleasure, thanks for watching!
How may 4 digit numbers can be formed from 1,2,2,3,4,5,6,7 without repetition?Can you give a solution asap
Can upload Vedio about 6 digit number between 0-9
what about if 3 or 2 digits are always the same
I have a 4 digit code I’m trying to figure out. I know the last two digits _ _ 6 9. Numbers 0-9 and there’s no repetition. How can I figure it @Wrath of Math
Thanks for watching Valeria and the question you asked has many solutions. Are there perhaps some details you've left out? If we know it is a 4-digit code, the last two digits are 69, and no repetition is allowed, then 1269, 1369, 4369, are all examples of valid codes, and there are many more. Are we trying to count the number of options? Is that the question?
Awesome
Thank you!
My head hurts from all this MATH
Thanks for watching and I'm sorry to hear of your physical discomfort! if I can be of any help to clear anything up and relieve the pain please let me know! Keep up the hard work!
Wrath of Math thanks I am ok
I mean thanks I am ok
I just want to unlock a 4 combination key lock, ending up learning math
You spilt some Bud Light Seltzer on yourself huh? Nice, lets learn! lmao
Haha, I don't drink so it was pure water and CO2, I promise! I like seeing my decorated math set in this video! Kinda makes me wanna decorate my set again for the season...maybe something spooky 😉 Thanks for watching!
@@WrathofMath those are great ideas. I did like the Christmas theme you had. Looking forward to what you come up with next! :)
The trailer has dropped! th-cam.com/video/9qtPX5U7r7o/w-d-xo.html
@@WrathofMath Loved the trailer! Very creative! You got the sub!
If we use 0 in ist digit place it will become 3 digit number ( because of insignificant) so there should be 9 possibilities in ist digit place i think?🤔
Thanks for watching, I'll copy a reply I made on this video to a similar comment. It entirely depends on the context of the problem. That's why this video has the word "codes" in the title. You might say 0123 isn't really a four digit number, but it certainly is a four digit code, as you might use for a lock combination. Does that help? If you wanted to disallow zero you'd have to specify that in the problem.
@@WrathofMath oh ok ok thanks🙌👀🤍
How about 4 digits, no repetition and the number is even?
how about any four digits? 4 distinct digits?
My problem is a bit different. I've got letters I, R, N and P and I've got to figure out all the combinations but without repetitions so I-I-I-R and R-I-I-I counts as one as they are the technically the same. I've computed 34 combinations but I don't know if I've missed any. I get a bit lost trying to use a brute force algorithm and I need to ensure I've got all the combinations. I had to do this for a 2 code length and 3 code length also. The 2 code length had 10 combinations and the 3 code one had 19 so I feel like I might have missed one or more there also. Anyone that could help, it would be greatly appreciated. Thanks.
Thanks for watching Alison! Could you clarify your question? You say "without repetition" but then list I-I-I-R as an option, which repeats I three times. If you're making combinations of four letters from I R N and P, without repetition, then there is only one option, which is to include all four of the letters. I imagine I am misunderstanding your question.
@@WrathofMath It's okay, I think 34 combinations is right. Bascially there can be repetition of the value but not the overall combination. So like your in video, a code of 9900 is fine but a code of 0099 would class as repetition as they contain the same numbers just in a different order. Using your method, there was 256 combinations but many of those will have repetition within them. So 9900 is fine but then I can't include 9009 or 0099 or 0990 because they each contain the same contents. I don't know if that helps? So for the two code length, I had [I-I, I-R, I-N, R-R, R-N, I-S, N-N, R-S, S-N and S-S] I can't also have [R-I, N-I, N-R, S-I, S-R and N-S] because they are the same as previous values, albeit in a different order. Hope that makes more sense? Tried a different method for brute force and got the same 34 combinations so I presume that is right but wondered if there was a maths trick to finding this out for certain, in case I have to increase to 5, 6, 7 and other code lengths in future. Thanks
Thanks for making this but i hope you'll include some condition like less than 7,000 hehehehe I'm really having a hard time understanding permutation
Thanks for watching Nicole, and that sounds a like a fun restriction to mix up the problem. I'll try to do a video on that soon. Think about how forcing the number to be less than 7000 restricts our choices of digits!
@@WrathofMath thank youuuu
This video isn't published yet, but with this link you can watch it early! th-cam.com/video/eoy8fHiWomU/w-d-xo.html
EDIT: HD version of the video hasn't processed yet, so if you watch it shortly after I post this comment its resolution will be a bit low
5 diget code with numbers 0158
Last number of the code is 0
How come we use digit 0 can be used in thousands place then it will be 3 digit number
Ex 0987
Thanks for watching and that's a great question! If we were strictly making numbers by selecting digits, then that would be a very important detail for us to point out! We would only have 9 choices for the first digit, because if - for example - we are trying to make 4-digit numbers, if we start with 0 it is really a 3 digit number. In this video though we weren't talking about numbers specifically, we were just talking about 4-digit codes, like you might use for a password or PIN - where place value doesn't really exist. Does that clear it up?
What about credit card numbers, can anyone have the same 16 digit number because i entered the wrong security code and that shit went through 💀
Zero is not included please
Sir repeat
Repeat what?
120
I was right
Although logically it is correct to use 0 as our first number, it isn't really the case. Most of the time 0 is not counted as the start of a 2 (or more) digit number so therefore in the case of the 4-digit number 0 can't be used on the first slot so that means there are only 9 out of 10 possibilities.
Thanks for watching and I would agree if I had framed the problem as "4-digit numbers", but I specifically framed it as "4-digit codes" because codes can start with anything, the sort you might set as your phone's passcode. But your point is an important one to keep in mind, sometimes we will be counting numbers - not codes - and then we do have to exclude 0 from the first digit's options!
Happy cirtmase
Thank you, it's coming up again soon!
10000
1234567890 number multiply 1234567890 which 4 number could not repeat in every last 10 number
It’s not 10,000
mggjkjjbkrkhkkhkhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhjh
One of the many fine examples of possible sequences of 298 characters from the 26 letter alphabet!
If you declare with your mouth, Jesus is Lord and believe in your heart that God raised him from the dead, you will be saved Romans10:9,Repent and Accept Jesus Christ as your Lord and Savior before its too late,or there will be consequences
u r asm
Thank you! I try my best! :)
@@WrathofMath sir why dont u make a video on limits and derivative
Thanks for watching, Siddha, and good question! I answer your question in this new video: th-cam.com/video/lP4kTCDB58E/w-d-xo.html around the 44:38 mark
Anjink
🤢🤢🤢🤮🤮🤮🤮
Thanks for watching and do you have any questions? Or are you just sick to your stomach?
@@WrathofMath 😠😠
😡😡😡
@@WrathofMath 😂