Britley Unilongo asked about how I am addressing the vowels at 14:39. Hi Britley, actually where I teach, in a question like this, students are expected to treat letters as if they were on scrabbled tiles. So letters, once again in a question like this, cannot be repeated. This is part of the curriculum I teach. The expectation is the for questions of this sort the letters cannot be repeated, while at the same time if the question was asking students to build a number out o available digits, they must assume that the numbers CAN be repeated unless they are told otherwise. So this conversation happens at the beginning of the teaching of this chapter: Letters cannot be repeated unless told otherwise, and numbers can be repeated unless told otherwise. If that looks really weird to you please talk to your teacher and find out what specific assumptions you are expected to make when you address these types of problems to clarify your question here.
Thank you for this video! I like the fact that you did not use any of the formulas; using only reasoning to set up the problems and get them solved. If I may ask, would it be out of bounds to ask for the PDF copy of the questions used? I think having those questions and following your presentation would be very helpful. Thank you once again for your effort; this is the best I have seen on this topic.
Hi Jeremy, Could you please show a Tree Diagram of the Last question (0,2,3,4) Digits ? I'm trying to learn how it becomes 49 using the Tree Diagram. Thanks !
Hello, I'm having a really difficult time understanding why there's 2 cases when solving for digits with even numbers. If you still place 5 at the end, and then 9 at the first digit, you would still be able to reduce the rest of the digits then multiply. I'm still confused on how it has to be 2 cases.
I'm not placing 5 at the end. When I put a 5 there I am saying there are 5 options for that digit which makes sense because there arw 5 even digits. The problem is that your choice of digit in the last position affects the number of choices you have for digit in the first position. If the digit you actually use at the end is a zero then you have all 9 other digits to choose from for the first digit. But if that last digit turns out to be anything but zero then you only have eight options for the first digit. This is because you can't repeat digits so you can't use the digit you already used and you can't use the zero. So because the choice of final digit affects the number of choices for another position we need to split this into two cases.
This is a collection from a lot of different people. Usually any textbook covering this topic will have a lot of examples. They are usually pretty easy to create.
If you are asking how many ways this can be done we use the fundamental counting principle and we start with the final digit. There are 5 options for this second digit. For the first digit there is only one digit we cannot use from the list given and that is the digit going in the second position. This means we have 8 options for that digit. When we multiply we get (8)(5)=40 different two digit odd numbers. Does that help?
I have a question that I am stuck on. My teacher had put down "(7!)-(6!2!)" for the answer and it just does not make sense to me on how he got that. Here's the question being answered: "7 different types of cars are being parked side-by-side. How many ways can they be parked so the convertible is not next to the subcompact?"
So 7! Is the total number of permutations of all the vehicles. Then what we do is find the number of permutations where those two vehicles are together. If we treat them as one unit then there are 6 objects to permute. Those six are the five other vehicles together with the two that we are keeping together as one unit. Then there are 2 factorial ways of permitting those two vehicles together. This gives is 6!2!. Now we subtract the ways they can be together from the total number of permutations and what we get is all the different ways they can be separate from each other. I hope that makes sense
@@JeremyKlassenThePiMan Ohh, thank you so much! Our teacher left us to teach ourselves this entire unit, so it’s a bit challenging being on my own lol. I appreciate your help!
From 5 Nepalese ,,4 Americans and 6 British ,a committee of 6 is to be formed .In how many ways this be formed when the committee contains exactly 2 Americans ??? How can we solve this question ????plzz upload the video for this question
Before I upload a video for it let me try this. First note that since there are no special positions on this committee that order is not important so we will be using combinations, not permutations. Since we need exactly 2 Americans there are 4C2 ways this can be done. Now we still need 4 more people but there is no other restriction indicated so we simply consider the entire group of Nepalese and Britains, a group of 11. We can choose 4 from this group in 11C4 different ways. We use the fundamental counting principle to determine the number of ways of choosing the whole group when we multiply these results together. The final answer will be (4C2)(11C4)=1980 different committees. I hope this helps. This may be a bit confusing if you are still working with just the fundamental counting principle and have not yet discussed permutations and combinations. Please let me know.
Okay. I will try, but first let me know. Have you started working with permutations and combinations or are you only working with the fundamental counting principle?
What if the question is: In how many ways can yiu place 9 different books on a shelf is there is soace enough for only 5 books? What can be an answer guys?
Well in that case you won't be able to put all the books on the shelf. You will only be able to put 5 of them. So the question then is, which 5 and in what order. This is a permutation in the form 9P5. This amounts to 9*8*7*6*5=15120.
Wow I'm watching and spend mostly 2 hours a day on a chapter that's only gonna be less then 10 Mark's in my upcoming test .And watching 30 minute videos 😂😂 Writing a Maths test tomorrow 😬👀🙌 wish me goodluck
You are good teacher. Keep it up. I enjoyed your lesson.
Britley Unilongo asked about how I am addressing the vowels at 14:39. Hi Britley, actually where I teach, in a question like this, students are expected to treat letters as if they were on scrabbled tiles. So letters, once again in a question like this, cannot be repeated. This is part of the curriculum I teach. The expectation is the for questions of this sort the letters cannot be repeated, while at the same time if the question was asking students to build a number out o available digits, they must assume that the numbers CAN be repeated unless they are told otherwise. So this conversation happens at the beginning of the teaching of this chapter: Letters cannot be repeated unless told otherwise, and numbers can be repeated unless told otherwise. If that looks really weird to you please talk to your teacher and find out what specific assumptions you are expected to make when you address these types of problems to clarify your question here.
Thank you for this video! I like the fact that you did not use any of the formulas; using only reasoning to set up the problems and get them solved. If I may ask, would it be out of bounds to ask for the PDF copy of the questions used? I think having those questions and following your presentation would be very helpful. Thank you once again for your effort; this is the best I have seen on this topic.
so helpful. i`m in my second semester,first year.it really help me alot.thank u
Thank you for helping me understand the counting principle
I liked the idea of showing off different exercises.. :), thank you :)
I still don't understand the solution for example 3 - 7:30
Hi Jeremy, Could you please show a Tree Diagram of the Last question (0,2,3,4) Digits ? I'm trying to learn how it becomes 49 using the Tree Diagram. Thanks !
Does this help. th-cam.com/video/ilxhBdoBlkE/w-d-xo.html
Hello, I'm having a really difficult time understanding why there's 2 cases when solving for digits with even numbers.
If you still place 5 at the end, and then 9 at the first digit, you would still be able to reduce the rest of the digits then multiply. I'm still confused on how it has to be 2 cases.
I'm not placing 5 at the end. When I put a 5 there I am saying there are 5 options for that digit which makes sense because there arw 5 even digits. The problem is that your choice of digit in the last position affects the number of choices you have for digit in the first position. If the digit you actually use at the end is a zero then you have all 9 other digits to choose from for the first digit. But if that last digit turns out to be anything but zero then you only have eight options for the first digit. This is because you can't repeat digits so you can't use the digit you already used and you can't use the zero. So because the choice of final digit affects the number of choices for another position we need to split this into two cases.
@@JeremyKlassenThePiMan Ahhh! I see! Yes this makes more sense to me now. Thank you very much!
Great 👍
0 itself can be a digit? For the last question
Yes
That's cool. From which textbook, you are solving these problems?
This is a collection from a lot of different people. Usually any textbook covering this topic will have a lot of examples. They are usually pretty easy to create.
What if two digit odd numbers can be formed using the digits 12345678 and 9,without repetition?
If you are asking how many ways this can be done we use the fundamental counting principle and we start with the final digit. There are 5 options for this second digit. For the first digit there is only one digit we cannot use from the list given and that is the digit going in the second position. This means we have 8 options for that digit. When we multiply we get (8)(5)=40 different two digit odd numbers. Does that help?
@@JeremyKlassenThePiMan perfect
I have a question that I am stuck on. My teacher had put down "(7!)-(6!2!)" for the answer and it just does not make sense to me on how he got that. Here's the question being answered: "7 different types of cars are being parked side-by-side. How many ways can they be parked so the convertible is not next to the subcompact?"
So 7! Is the total number of permutations of all the vehicles. Then what we do is find the number of permutations where those two vehicles are together. If we treat them as one unit then there are 6 objects to permute. Those six are the five other vehicles together with the two that we are keeping together as one unit. Then there are 2 factorial ways of permitting those two vehicles together. This gives is 6!2!. Now we subtract the ways they can be together from the total number of permutations and what we get is all the different ways they can be separate from each other. I hope that makes sense
@@JeremyKlassenThePiMan Ohh, thank you so much!
Our teacher left us to teach ourselves this entire unit, so it’s a bit challenging being on my own lol. I appreciate your help!
that was helpful thanks alot!
From 5 Nepalese ,,4 Americans and 6 British ,a committee of 6 is to be formed .In how many ways this be formed when the committee contains exactly 2 Americans ???
How can we solve this question ????plzz upload the video for this question
Before I upload a video for it let me try this. First note that since there are no special positions on this committee that order is not important so we will be using combinations, not permutations. Since we need exactly 2 Americans there are 4C2 ways this can be done. Now we still need 4 more people but there is no other restriction indicated so we simply consider the entire group of Nepalese and Britains, a group of 11. We can choose 4 from this group in 11C4 different ways. We use the fundamental counting principle to determine the number of ways of choosing the whole group when we multiply these results together. The final answer will be (4C2)(11C4)=1980 different committees. I hope this helps. This may be a bit confusing if you are still working with just the fundamental counting principle and have not yet discussed permutations and combinations. Please let me know.
I have so much confusion with this question ,,so if u provide the solution then it will be helpful for me ,,so plzzz
Okay. I will try, but first let me know. Have you started working with permutations and combinations or are you only working with the fundamental counting principle?
Have a look at this. th-cam.com/video/WausjdGYqIA/w-d-xo.html
Thank u! We have an assignment yoday due 9 PM and omg i didn't know what to do HAHHAHA
What if the question is:
In how many ways can yiu place 9 different books on a shelf is there is soace enough for only 5 books?
What can be an answer guys?
Well in that case you won't be able to put all the books on the shelf. You will only be able to put 5 of them. So the question then is, which 5 and in what order. This is a permutation in the form 9P5. This amounts to 9*8*7*6*5=15120.
im wondering....why prioritze the 9 8 7 6 and 5? shouldnt it be 1 2 3 4 5?
Permutation formula. Try it on scientific calculator. Input : 9 and then click SHIFT, click nPr sign, input 5.
9P5=15,120
thank you very much sir
Thanks a lot! :D
Thanks!
20:59 no repeititions
I'm not sure I understand what you are referring to. The answer given in the video did not repeat the digits.
@@JeremyKlassenThePiMan oops I got the time wrong
@@bigidlagger7589 I think the question at that time stamp also doesn't include repetition.
@@JeremyKlassenThePiMan really? oh
@@bigidlagger7589 at least I am pretty sure
Wow I'm watching and spend mostly 2 hours a day on a chapter that's only gonna be less then 10 Mark's in my upcoming test .And watching 30 minute videos 😂😂
Writing a Maths test tomorrow 😬👀🙌 wish me goodluck
good luck on the math test. I'm doing upgrading to so days id rather clean my house then do math.