24:43 I don't really agree with what you said here. The two letters can not only be 1st and 2nd (etc.) but also 2nd and 1st. If the two letters were switched, it would count as a different arrangement. Unless those letters were the same. So, wouldn't there be more to consider?
I enjoyed refreshing myself on the different ways of efficiently counting possibilities, including stars and bars (which I initially wasn't familiar with)!
Great video, on the last problem, a different approach would be to notice that one-half of the time, when Laurent's number is between 2017 and 4034 it is always greater than Cloe's. Now for the case that they choose a number from the same interval, we find that the probability that L's is bigger than C's is 1/2. 1/2*1/2+1/2=3/4.
This is great help. But for 2003 AMC 12A -- how can I know if they want the combination or permutation? "Arrangement" makes it seem like permutation, which would be 2^15. Is "arrangement" always understood to be a combo?
For 9:26, should the answer be 67 not 72? I got 67 before and when I checked your steps, I got confused. So I went to google I went to search up the answers and it came up as 67. Even the AOPS website stated it themself.
I have a question, for the license plate problem 2 letters and 4 numbers, don’t we have to multiply four factorial because of you have the letters on the first and the second, there are still 4! Ways to arrange the other four number on the third, fourth, fifth, and sixth.
That's a good question. We don't actually multiply by 4! because we are arranging the numbers already: our 10^4 represents any possible 4 number combination, already accounting for order.
![Summer Walker - Heart Of A Woman [Official Lyric Video]](http://i.ytimg.com/vi/SFddctbZzw8/mqdefault.jpg)
I know I'm late but this video series is incredible and worth paying for. Great job.
Glad you enjoy it!
24:43 I don't really agree with what you said here. The two letters can not only be 1st and 2nd (etc.) but also 2nd and 1st. If the two letters were switched, it would count as a different arrangement. Unless those letters were the same. So, wouldn't there be more to consider?
I especially enjoyed the geometric probability segment, thanks!
I enjoyed refreshing myself on the different ways of efficiently counting possibilities, including stars and bars (which I initially wasn't familiar with)!
Yup, stars and bars is definitely on the harder end of the AMC 10, but it can help make problems a lot easier!
I enjoyed learning about how to use casework cleverly to solve the problems faster but also in a comprehendable way
That's great to hear!
This helped restore my lacking combo confidence, thanks!
Great to hear!
I liked that there were real AMC problems used to explain the concepts.
Yup, we think real AMC problems provide the best practice!
I learned about stars and bars, very helpful video!
I am so hyped for the class hope you can start now. LOL!
Also please pick me for the giveaway. LOOOOOOL!
yay!
After watching this video my Counting and Probability is so good. I used to be so bad at Counting and Probility. Thanks Puzzlr.
Of course, you're welcome! That was our goal!
Also enjoyed the clarity in which you solved the questions!
Great video, on the last problem, a different approach would be to notice that one-half of the time, when Laurent's number is between 2017 and 4034 it is always greater than Cloe's. Now for the case that they choose a number from the same interval, we find that the probability that L's is bigger than C's is 1/2. 1/2*1/2+1/2=3/4.
Thank you for the good tip on how to salve combinatoric questions
Of course, you're welcome!
Please make more of these videos. I think that they are helpful! I’m trying to improve on the AMC 10 myself 😄
agree!!!!!!!!!!!!!!1
Thank you! I liked how well you explain the concepts you use, and how to use them to solve problems.
Of course, you're welcome!
I really enjoyed the clarity through which you went through the problems. It really made it extremely easy to understand the concepts.
Of course, thanks for the compliment!
This is great help. But for 2003 AMC 12A -- how can I know if they want the combination or permutation? "Arrangement" makes it seem like permutation, which would be 2^15. Is "arrangement" always understood to be a combo?
This man seems very smart
Why thank you! Hope this video helped you on your mathematical journey!
This class helped me a lot.
Comment below if you're hyped for class!
is it normal to face a lot of difficulty even in the beginning? i'm an incoming 9th grader
Yes! It takes time & practice, but keep at it, and you'll qualify for AIME in no time!
wonderful upload ThePuzzlr. I killed that thumbs up on your video. Always keep up the high quality work.
Hi, I'm so excited for class today!
Me too!
Hi Puzzlr uhh for the pascal’s triangle, in the last row it should be 84+126=210 not 84+126=200 but it is a great video =D
This class helped a lot! Thanks!
Of course, you're welcome!
how to tell if order matters or doesn't matter in a counting problem, for the combinations and permutations problems
For 9:26, should the answer be 67 not 72? I got 67 before and when I checked your steps, I got confused. So I went to google I went to search up the answers and it came up as 67. Even the AOPS website stated it themself.
yea he got it wrong
I liked the applications of the problems
Yup, we tried to show real application of the strategies so that you can best apply it on REAL AMC problems!
i dont understand any of this, what shud i do or am I just dumb
A small correction over there? 50 + 33 = 83 rather than 88
at 24:35 how do you know there is 10 choices for each letter?
"casework should only be used as a last resort"
meanwhile 99% of amc 10 probability questions: casework go brrr
lol
at 9:37 I think you miscounted 50 + 33 to be 88 instead of 83
Yup, sorry about that. It should be 83 not 88.
8:45, just a careless mistake, but the video was a great review, thanks
Good call!
I have a question, for the license plate problem 2 letters and 4 numbers, don’t we have to multiply four factorial because of you have the letters on the first and the second, there are still 4! Ways to arrange the other four number on the third, fourth, fifth, and sixth.
That's a good question. We don't actually multiply by 4! because we are arranging the numbers already: our 10^4 represents any possible 4 number combination, already accounting for order.
Very Helpful!
Glad you thought so!
Very good video
Thanks!
Quick correction 50+33 is 83 i think or im insane but anyways this video is worth it lol
at 9:05 isnt the sum 83 and the answer 67
Yup, sorry about that. It should be 83 not 88.
25:00
theres a mistake at 8:59 50+33 doesnt equal 88 XD
For overcounting, 50+33 is 83 dumbo. The answer is 67