AMC 10 Skills: Stars and Bars (5 Examples)
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- Опубликовано: 24 окт 2024
- Time Stamps and Practice Problems Below.
website if interested in classes: thebeautyofmat...
These 5 examples will also appear on the 2016 AMC 10 A Problem 20 video. But I wanted to have them appear in a stand alone video as well, for those who just wish to learn this particular concept as a pursuit in and of itself. Example 2 5:20, Example 3 9:12, Example 4 11:34, Example 5 13:16 2016 AMC 10 Problem 20 queued to the right moment: • 2016 AMC 10 A, Problem...
For AMC level Practice, try these problems: artofproblemso...
try not to see the solution, maybe even search the problem up without clicking the link(link will be problem with solution right under it)
Pretty good teaching! It explains the logic so well rather than jus memorizing the formula
Thank you for the compliment. I was definitely proud of this one, but more importantly, learning how to understand it to be able to explain it was a very enjoyable process.
As a person who was extremely new to this concept, I jumped from video to video but I've never seen anything as good as this. Thank you so much for creating this!
Thank you! Glad to hear that the video fulfilled it's objective. Good luck in your educational journey!
This is the best stick and stones (stars and bars whatever you call it) video I’ve ever seen.
Thank you Christian. It means a lot. :)
The only video on bars and stars that enlightened me with some insight with very simple explanation
Thanks Sagar...glad it was a helpful experience for you. Best of luck!
Thanks bro. I understood all examples 👍🏻👍🏻👍🏻 and concepts
This is the best video I’ve ever seen on stars and bars!! Thank you so much :)
Thank you so much for the feedback! Good luck in your preparation!
I was trying to find a good explanation of this concept for my child and this was the best one I have found! I also appreciate the fact that you don't rely on a formula but instead reason out what you are counting. Thank you!
Thank you so much for saying so Jamie. It really helps to know the impact the videos are having and how it is able to help others. Is your child preparing for one of the AMC tests? If so, and if you don't mind sharing my I ask the grade level? I have a great video for development planning I want to recommend if you are interested. It is this one:
ruclips.net/video/OD3f_tqN-OE/видео.html
You teach your kid discrete math? may i ask why (very interesting) ?
Hello, I've just stumbled across this video since I was struggling with a few questions. I don't think I've commented anywhere in the last couple years but I just wanted to say that the first few examples you went through helped me tremendously with my work and I'm really grateful.
Thank you so much for sharing such feedback. I am equally grateful to know the content was useful to you. Are you preparing for school material or for a Competition?
Beautifully explained , Thankyou!!!!!
holy shit this is gold. cramming before my exams and put off perms and coms. Teacher didn't teach this method and I found the m+r-1 etc. formulas online without context yet this clarifies them so well!
Thanks Kita, what class are you taking? Combinatorics?
@@TheBeautyofMath I'm Taking High Level Mathematics in the IB diploma program. Exams are in a few weeks time!
i stubled upon this video becuase i was having trouble with the stars and bars problems on amc 10. thank you for helping
No problem! Glad it helped!
You should definitely teach more concepts like these!!! This video provides a solid introduction/understanding of stars and bars in under 20 min, which not many people are able to do. :)
Also one quick thing, in the amc's how are we able to recognize problems that use stars and bars to solve?
Thanks. Yeah, I will do a few more like this, but I am under contract to develop a teaching series over fundamental concepts on the AMC 10 and if I do the same content for free on my channel I will he violating the contract, however I can still do some topics, and even eventually share some of that video content here. But it's still in development phase.
you are great man
These skill videos help so so much. Thank you for this, is it possible for you to eventually create other AMC 10 skill videos.
Thats the plan. :) summer time is hard though. 50 hour work weeks. 4 more weeks left. Then I will start pumping out a lot more content rapidly. Gonna go film later today. The Skill videos require more time consumption as I have to gather problems from various sources to make a quality video. The AMC 10 content requires much less time and so I can film it easier. But it will get done. :)
Ishaan I changed my mind. I will try and film 1 skills video a week in the summer starting next Saturday will be first film day, can you make a list of 10 to 20 suggested topics for the videos in a comment? I will maybe make a video requesting it from all subscribers as well. I don't know every topic, for instance I need more time to research Chinese Remainder Theorem before I feel I can explain it comfortably, that I will research later, but name some topics you would like to see, and I will begin structuring them. Thanks!
@@TheBeautyofMath Thank you so much for doing this.
Some Possible topics could be:
Modular Arithmetic
Binomial Theorem
Pascal's triangle
Hockey Stick and other binomial identities
Perhaps some circle stuff (arcs, cicrumference, sectors, tangents, etc.)
Power of Point
Trignometry
Quadratics (vietas, discriminant, differnce of squares, etc)
Different types of triangles and their properties
Angle Bisectors, medians, altitudes
Constructive counting
Casework counting
Fermat's Little Theorem
Sequences and Series
PIE
Thats all I can think about right now, thank you again :)
Ishaan I am sorry. Can't do this week. In order to make the videos I need to assemble about 5 questions or so depending on the topic that illustrate the concept being taught.
But I just haven't had time. I need to go through like 20 years of AMC's and break them down by underlying topic to do that, then go through AoPS books and find other relevant problems if there is not enough content hidden inside of the AMC's. The teaching part isn't the hard part, it's assembling the right group of 5 or so problems. I will try and work on it this week. So today just going to film normal content.
Hi, I'm confused by something at 3:40. I understand that 6C2 works, but I also feel like is 25. This is because, if each friend has at least one candy, the problem basically simplifies to splitting 4 pieces of candy between friends, with no restrictions. In order to solve this problem, we can place the 2 bars in any of the five spaces, resulting in 5 squared. So what did I do wrong?
When you want to split 4 pieces of candy between the three friends, you have 4 pieces of candy and two bars. You can then ignore the 'at least one' constraint and do the regular stars and bars formula which is (s+b)!/b!s! where s is the number of stars and b is the number of bars. In doing this, you get 6!/4!*2 which is essentially 6C2.
I'm not sure I quite understand how you got 5 spaces, but feel free to reply and further explain your reasoning. :)
HI Ray. Great question. Sorry for the delay in response. Just been too busy. It is basically because the way you are attempting it isn't actually rendering the problem correctly. If you place 4 C's for the 4 remaining candies after they each get one: CCCC and look at the 5 spaces, you might think that there are only 5 choices in that set up, but in fact every space has 2 positions for Dividers. So there are actually 10 spaces where we could put the 2 dividers. But that would be 10C2 which clearly isn't going to work and that is probably because of over-counting. This is why we typically don't approach that track of solving in that manner. Instead We call the Divider's D's. we place all 6 letters down: CCCCDD. Then we ask how many positions can the D's go in? Since there are 6 letters, It again converts to 6C2. This is the same method used in Example 2 if you watch that part of the video. I talk about exactly what you are mentioning and then show the above letters method. I can't really explain what the over counting is caused by or how it could be corrected, I just recognize that it isn't going to work and realize I have to use the letters method instead. I probably learned why it doesn't work at some point, but such thoughts have been lost to memory holes long ago. Let me know if Example 2 from this video and my above explanation helps. Thanks. Upon further thought, I can see that one of the reasons it overcounts is with the 2 spaces for Dividers in each "space" it would look like this _ _C_ _C_ _C_ _C_ _. But let's say you chose space 1 and space 3. now it would look like this D_CD_C_ _C_ _C_ _. where by 1 and 3 I mean the little _. This would mean person A got 0 B got 1 and C got 3 of the remaining 4 candies. Now what if you placed the dividers in spaces 1 and 4. Now it would look like this: D_C_DC_ _C_ _C_ _. But that is actually the same 0/1/3 outcome. Even though we have counted it as a different set up. This is why the Example 2 method must be used.
@@nolanyeemusic the spaces he means around the 4 candies are the space in front of the first candy. The 3 spaces between the Candies, and the space after the last candy for a total of 5.
I have a question about 4:32. Why can't we solve the problem by giving each friend once piece of candy so that by default each friend has one piece. Then we have 4 pieces of candy remaining and we have to figure out a way to divide this amongst 3 friends, so is it not just 4 choose 3 which is 4? I know this doesn't make sense numerically but where is my logical mistake?
First you can approach it as each friend gets one piece of candy right off the bat. Then indeed you have 4 pieces of candy remaining and we have to figure out a way to divide this amongst 3 friends where 0 is an allowable amount. The next part is your mistake. 4 choose 3 does not work here. It is actually nonsensical. you are choosing 4 pieces of candy from 3 friends. Friends are in fact not candy. *a joke* When you are using combinations the things you are choosing must be equivalent to what you are choosing from. Instead what you would need to do has several approaches. one would be case work. you could say case 1 is 4/0/0. where one person gets all 4 and the others get 0. This can happen in 3 ways. Case 2 could be 3/1/0 which is 3!=6 ways. Case 3 2/2/0 which has 3 ways. and Case 4 2/1/1 which has 3 ways. Add these up and you get 6+3+3+3=15. The same as 6 choose 2. Another could be a variation of stars and bars where you say CCCCDD, how many arrangements? The D's are dividers. The C's are Candy. For example the CCCCDD Would be equivalent to 4/0/0. Let me know if this is understood now.
Wow, I just realized how flawed my solution was lol. Thanks for the help.
@@TheBeautyofMath I also had another question. With the AMC10A being on this this Thursday, from your personal experience what would you recommend I do 1 day before the test and on the morning of the test? Should I take it easy the day before or try and jog my memory as much as possible? On the morning of the test should I wake up 1-2 hours before and warm up my brain to start thinking?
Video coming out around 1130pm tonight Pacific time discussing exactly that. You can watch it early tomorrow.
@@TheBeautyofMath Thanks!
when are you supposed to count the number of stars and bars together and when are you supposed to count the number of gaps?
It is kind of intuited. But in general, if you are allowed to assign 0 as a value to an individual then you would not use the gaps method. The gaps method relies on no more than one divider per gap. Making 0 impossible. However when it says "each person receives at least one" that is when you use the gaps method for precisely the reason above.
@@TheBeautyofMath If you are allowed to assign 0 then you would count the number of stars and bars, correct?
Precisely.
Qnd if someone of them can have 0 candy !? Thanks
I think you found this example later in the video? :)
Hey tbom, I have a question
Let’s say there are 20 candies and u have 3 ppl, how many ways are there to distribute the candy (u can give everyone 0 candy)
So I wrote c c c c … 20 times and counted 21 spaces from the left of the first c to the right of the last c, so wouldn’t the answer just be 21c2?
Also from the video, isn’t DDCCCCCC the same as DCCCCCD since both times involve a person getting 2 candies
This is a good thought. I will answer with a question. See if it reveals the fault in your current understanding: how would the middle person get 0 candies?
Also just saw the other comment in the filter. I am not sure you understand what DDCCCCCC notation means. D=Divider, C=Candy. So the first option you listed is person A/B/C gets 0/0/6 respectively. The second one is person A/B/C getting 0/6/0. In neither case you listed does anyone get 2 candies. There are 2 people getting 0 though.
People are unique. So just because in both scenarios 2 people get 0 candies, it isn't the same because they are different people getting 0. Think of it like this let's say between you and 2 other people(3 people) that 2 of you would get 1 million dollars $. Would it matter to you which 2 received it?
@@TheBeautyofMath idk
@@shangjiang3320 ok so assuming the "idk" is in relation to the first question of how would the middle person get 0 candies, you are correct in saying "idk" as there isn't really a way to do it with that method of gap choosing. That's why you need to use the C's and D's method. So for the problem you gave with 20 candies and 3 people you need 20 C's(Candies) and 2 D's(Dividers) then you have 22 Letters and you need to choose 2 spaces to put the Dividers(D's) in. 20 the calculation is 22 Choose 2
EXCELLENT
I still don't quite understand the 6 choose 2 part. I understand the 6, but why is there a 2? Thanks
Which time stamp in the video? I don't remember the context. The time stamp will help so I can explain. Thanks.
@@TheBeautyofMath 3:40-4:10
@@masteryoda5237 because there are 6 gaps. You are making 3 nonzero sized groups. In order to make 3 groups, you need 2 dividers. That's where the 2 is from. The 2 dividers. For instance. CCdCdCCCC would be like the 6 choose 2. The first person gets the 2 C's. Second person gets 1 C. Third person gets 4 C's. But it only take 2 dividers to make the 3 groups.
@@TheBeautyofMath Oh i see. Thanks for your detailed explanation.
@@TheBeautyofMath you're basically *choosing* 2 places to place the bars
Amazing Video
U R AMAZING
Thank you, friend. Glad it helped you in your understanding.
thankyouu
Always welcome. Good luck in your studies.
hahahah it was on the 12a question 18 or something
@@alankuo2727 hey just curious did you find a stars and bars technique that worked for question 18 on AMC 12A fall? I was unable to and used a different process. Please do share if you found the solution that way how you calculated it. If you put it on this video, I will pin the comment: ruclips.net/video/TOSHQPb7vaM/видео.html
@@TheBeautyofMath hey
stars and bars was the first thing I thought when I saw the question but I didn't successfully get an answer
now that I've looked at it more I don't think I have a way to do it with stars and bars unfortunately