your teaching methods and your techniques are so easy to understand and effective. this wasn’t really the feeling I got in paid tutoring. thank you so much!
BRUH this video taught me such an important concept i really appreciate this thank you so much I did the past exams and got like average of 23 with your channel this can be a 25 thank you so much continue making videos!
For q4 mrs jones - if one family brings dishes 1,2,3 and the second family brings dishes 4,5,6 doesn't this mean the third family can only bring dishes 7,8 which is only 2 rather than 3 dishes?
Hello, the question states that no two families can bring the same combinations of 2 dishes. For example, a family could bring 1,2,3 and another family could bring 1,2,4 and that would be ok since they don't have the same exact 3 dishes.
I'm a bit confused about 51:33 When would you have subtracted it 3 times? I don't know if this is a stupid question but...anyway thanks for the content! Just subbed :)
That is how PIE works. Let's focus on the common section where the three grandkids G1&G2&G3 all call. We add the days when G1 calls, G2 calls, G3 calls which includes that section 3 times, then subtract the days when G1&G2 call, the days when G1&G3 call, and the days when G2&G3 call, which subtracts that area 3 times. So we need to add that section back when G1, G2, G3 all call.
amc 2015 middle school 8 problem - since each team plays each other twice - then total number of games will be 112 because each team will play 7 other teams twice - which means each team will play 14 games. 8 teams times 14 games = 112 games SOOO final answer should be 112 plus 32 = 144!!! please check
Yes, each team is playing 7 other teams twice, but then we are doublecounting the games (A plays B, is the same as B plays A). That is why we need to do (8 choose 2) to get the team pairs for games, then multiply by 2 (since they play each other twice). (8 choose 2) = 28 x 2 = 56 total games within the conference. Now add 8 x 4 = 32 games outside the conference to get 88.
Just keep practicing and you will slowly build the intuition to solve complex problems. You can use the Mastering AMC8 and Mastering AMC 10/12 books and related videos to learn all the concepts you need.
You find the number of numbers that are multiples of 2, 3, 5 but that overcounting the multiples of 2 * 3, 3 * 5, and 2 * 5. But then, the multiple of 2, 3, and 5 are being counted 3 times in the beginning, subtracted 3 times, so know we are counting it 0 times but we must count it 1 time so we add the number of multiples of 2, 3, 5 back.
If the dish types were A, B, C, D, E etc, and if one family is bringing A, B, E, then no other family can bring ABE, but they can bring ABC, ABD, BCD, BCE, BDE. Hope it clarifies.
Hello, this is a great channel! One question though, (even though I'm two years late hehe) for example 2, why do we use combination? If each team played the others twice, shouldn't each team have played 14 games so therefore a total of 112 conference games?
Basically we need to find out how many pairs of teams are there, which is "8 choose 2" (which is 28). Then each team plays each other twice: 28 x 2 = 56. Each team also plays 4 games against non-conference opponents, which gives 8 x 4 = 32. So the total number of games are 56 + 32 = 88.
Hi, I am not the owner, but he is a teacher for the amc8 zoom class and while he was teaching it to his student, he kindly recorded it for people like us :)
There are 4 blue edges, 4 red edges, 4 orange edges which are parallel. Out of the 4 blue edges, any combination of 2 will be a pair of parallel edges. So how many pairs of blue parallel edges: "4 choose 2". Similarly "4 choose 2" pairs of red parallel edges, and "4 choose 2" pairs of orange parallel edges. You add them up to get the total number of pairs of parallel edges.
In the first example/introduction problem for casework at around 15:30 , you solved the case for if there were 0 5 dollar bills. But shouldn't we just ignore that because the questions ask a combination of 5 $ *and* 2$ bills. Ik that in this question it did not affect your final answer because the case didn't work, but in a question that needed a total of let say 18 dollars wouldn't it be wrong to consider a case with 0 5$ bills? Thanks for reading
A combination of $5 and $2 bills doesn't mean you have to include at least 1 $5 bill and at least 1 $2 bill. The problem would have specified that if that were the case.
@@shradhars9820 That's a pretty good score for first time. Anyone can take the AMC 10 contest and unrelated to AMC 8. So there is no qualification. You should learn AMC 10 material from the Mastering AMC 10/12 book and take the contest in November.
For the bottle problem, since the 2 T's and 1 O must remain together, just treat them as 1 group. Now you have 4 objects to order: B, the group, L, E so there are 4! = 24 ways to arrange them. Now, we also must arrange the elements in the group. This can be done in 3!/2! ways by the word rearrangement formula in class 1 because there are 3 letters (T, T, O) and 2 T's so we must divide by 2!. This is equal to 3 * 2!/2! = 3. So we then, our answer is just 24 * 3 = 72 because there are 24 ways to arrange the objects and 3 ways to arrange the letters inside the group.
your teaching methods and your techniques are so easy to understand and effective. this wasn’t really the feeling I got in paid tutoring. thank you so much!
Thanks, glad to hear that 🙂
@@SohilRathi you inspire me to do my best every single day.
Thank you so much for these videos ~ You explain very cleary!
this complementary counting was way too good. I have been doing AMC myself and finally I found a channel that substitutes for an institute
This helped me alot I was stuck on that question for a long time
BRUH this video taught me such an important concept i really appreciate this thank you so much I did the past exams and got like average of 23 with your channel this can be a 25 thank you so much continue making videos!
Bro how do you get 23. That is a very high score.
hello sohil rathi. im trying to prepare for a different competion. this is really helpful
😃
Best of luck!
For q4 mrs jones - if one family brings dishes 1,2,3 and the second family brings dishes 4,5,6 doesn't this mean the third family can only bring dishes 7,8 which is only 2 rather than 3 dishes?
Hello, the question states that no two families can bring the same combinations of 2 dishes. For example, a family could bring 1,2,3 and another family could bring 1,2,4 and that would be ok since they don't have the same exact 3 dishes.
Why do you divide by 2 factorial in question 6 of the homework. Can u please explain? Thanks
2 T's have 2 ways of being arranged, but they are the same arrangement. Also, you can see this by the word arrangement formula.
@@SohilRathi Oh, ok, thank you 🙏
Thank you so much for these AMC preparation videos! They are helping me a lot. I subscribed.
Glad you like them!
Thanks Sohil. You explained the concepts better than my college math teacher, no joke!
Glad to hear that!
Thank you so much!
You're going to go far in life. Great work!
Thanks!
fr he's like going to ivy
@@smaransure2234 lol
@@smaransure2234 It that you???
good channel i'm preparing for amc this channe's helpful
I'm a bit confused about 51:33 When would you have subtracted it 3 times? I don't know if this is a stupid question but...anyway thanks for the content! Just subbed :)
That is how PIE works. Let's focus on the common section where the three grandkids G1&G2&G3 all call. We add the days when G1 calls, G2 calls, G3 calls which includes that section 3 times, then subtract the days when G1&G2 call, the days when G1&G3 call, and the days when G2&G3 call, which subtracts that area 3 times. So we need to add that section back when G1, G2, G3 all call.
amc 2015 middle school 8 problem - since each team plays each other twice - then total number of games will be 112 because each team will play 7 other teams twice - which means each team will play 14 games. 8 teams times 14 games = 112 games SOOO final answer should be 112 plus 32 = 144!!! please check
Yes, each team is playing 7 other teams twice, but then we are doublecounting the games (A plays B, is the same as B plays A). That is why we need to do (8 choose 2) to get the team pairs for games, then multiply by 2 (since they play each other twice).
(8 choose 2) = 28 x 2 = 56 total games within the conference. Now add 8 x 4 = 32 games outside the conference to get 88.
In the mastering amc 8 book chapter 2 combinations numbers 2.4.2 and 2.4.5 are the same
Thanks for letting me know. I have removed the duplicate and it will be updated in next version of the book.
Can please give us any tips how to become pro in problem solving?
Just keep practicing and you will slowly build the intuition to solve complex problems. You can use the Mastering AMC8 and Mastering AMC 10/12 books and related videos to learn all the concepts you need.
Great video good job
Thanks! Glad you enjoyed it!
Just wanting some more clarification, why does "choosing" correlate to combination?
because if you choose 2 pencils out of 10 for example it's just the final subset of pencils that matter
For Example 10, I don't get how you solved it. Can you please explain?
You find the number of numbers that are multiples of 2, 3, 5 but that overcounting the multiples of 2 * 3, 3 * 5, and 2 * 5. But then, the multiple of 2, 3, and 5 are being counted 3 times in the beginning, subtracted 3 times, so know we are counting it 0 times but we must count it 1 time so we add the number of multiples of 2, 3, 5 back.
19:50 saving my lesson
And also in math class 4, in the first problem, how did you know that there were 36 choices
There are 6 choices for each dice, so the total number of choices is 6 x 6 = 36
Hi for the homework question number 4 I didn’t really get no two family bring the same combination of dishes
If the dish types were A, B, C, D, E etc, and if one family is bringing A, B, E, then no other family can bring ABE, but they can bring ABC, ABD, BCD, BCE, BDE. Hope it clarifies.
ohhhh ty i ust finished my amc 8 today@@SohilRathi
Is it possible to still sign up for your online lesson or is it too late? Btw you are a great teacher!
Yes, you can still sign up at omegalearn.org/amc8-advanced however 3 classes are already over so you can watch the recording of those.
Hey daniel how you doin'
You are really awesome
Thanks!
thx
is it possible to still sign up for your online sessions or is it too late?
There are no live classes right now, however you are welcome to sign up for future ones.
@@SohilRathi but how can we join the future classes?
@SohilRathi can I join now as well or is it too late 😢
you should've added chapters so i could have found the questions easier!
Hello, this is a great channel! One question though, (even though I'm two years late hehe) for example 2, why do we use combination? If each team played the others twice, shouldn't each team have played 14 games so therefore a total of 112 conference games?
In your method, each game is counted twice. The first teams plays and the 2nd team plays the same game. You are overcounting by a factor of 2
do you go to aops
Hey can you explain 18:23 because I am in six grade I don’t really understand😢.
Basically we need to find out how many pairs of teams are there, which is "8 choose 2" (which is 28). Then each team plays each other twice: 28 x 2 = 56.
Each team also plays 4 games against non-conference opponents, which gives 8 x 4 = 32. So the total number of games are 56 + 32 = 88.
41:50 88-16 is equal to 72
We are doing 83 - 16
Just wondering, is this a recording?
Hi, I am not the owner, but he is a teacher for the amc8 zoom class and while he was teaching it to his student, he kindly recorded it for people like us :)
Wait in 4:26 how did the 2 numbers cancel out??????
24/6 = 4, so you get a 4 in the denominator. Then 8/4 = 2, so you get a 2 in the numerator.
Ok thank you!
Hi can you explain 26:16 pls I am a bit confused.
There are 4 blue edges, 4 red edges, 4 orange edges which are parallel. Out of the 4 blue edges, any combination of 2 will be a pair of parallel edges.
So how many pairs of blue parallel edges: "4 choose 2".
Similarly "4 choose 2" pairs of red parallel edges,
and "4 choose 2" pairs of orange parallel edges.
You add them up to get the total number of pairs of parallel edges.
Thank you very much for replying and explaining. I understand now. ❤
4:20 Please I don't get why the 8, 6 and 24 cancel each other
8 x 6 = 48 divide that by 24, we get 2.
@@SohilRathi Thanks, I've understood it now.
In the first example/introduction problem for casework at around 15:30 , you solved the case for if there were 0 5 dollar bills. But shouldn't we just ignore that because the questions ask a combination of 5 $ *and* 2$ bills. Ik that in this question it did not affect your final answer because the case didn't work, but in a question that needed a total of let say 18 dollars wouldn't it be wrong to consider a case with 0 5$ bills?
Thanks for reading
A combination of $5 and $2 bills doesn't mean you have to include at least 1 $5 bill and at least 1 $2 bill. The problem would have specified that if that were the case.
@@SohilRathi Ok thanks
Just wondering, did you take the AMC 8? If so, what did you get?
I have taken the AMC 8 for the past 3 years and gotten honor roll when I was in 4th grade and distinguished honor in 5th and 6th grade.
@@SohilRathi Wow!
@@SohilRathiI only took it for the first time this year and got 19/25? Is that enough to qualify for amc 10?
@@shradhars9820 That's a pretty good score for first time. Anyone can take the AMC 10 contest and unrelated to AMC 8. So there is no qualification. You should learn AMC 10 material from the Mastering AMC 10/12 book and take the contest in November.
Wait a minute is the kid the teacher or not
Explain the bottle question
For the bottle problem, since the 2 T's and 1 O must remain together, just treat them as 1 group. Now you have 4 objects to order: B, the group, L, E so there are 4! = 24 ways to arrange them. Now, we also must arrange the elements in the group. This can be done in 3!/2! ways by the word rearrangement formula in class 1 because there are 3 letters (T, T, O) and 2 T's so we must divide by 2!. This is equal to 3 * 2!/2! = 3. So we then, our answer is just 24 * 3 = 72 because there are 24 ways to arrange the objects and 3 ways to arrange the letters inside the group.
@@SohilRathisir, why are you dividing by2 factorial? I don’t get it. Please respond ASAP. Thank you
46:55
What grade are you in
I am currently going to 7th grade
You are so good! Which state do you live in. I am going to 7th grade too. How and where do you learn all these topics? Thanks so much!
@@poonamchauhan1318 I am 12 and live in New Jersey in the US.
@@anirudhvenkatesan2230 I know where you live
AMC is australian math comp so he is in aus
I am a third grader why should I know?
you will need to know these things pretty soon so better get ready
Dude, how is he still at 481 subs
Now 717
@@waffles_1823 now its 795
@@redboo6863 Now it's 867
@@wenwenstudios6755 NOW ITS 1.91K
now it is 2.08K subs
hoiiiiii yaaaaaaaaa
56:50
33:25