Combinations Example 1: 4 Digits

[@varunnagp3903]

good lecture. I think students studying you are in good hands

[@NatashaSharma-c5k]

LOVE your lectures. your teaching method really is amazing and helps one see the beauty of math. You have done perfect justice to the topic. I wish all the math teachers were like you. I am a math lover in Grade 11th and would love if you could recommend some books that provide conceptual clarity in math(books that are not just a bunch of complicated math problems but really dig into the topic)

[@GamerMinecraftivity]

8:40 nice

[@user-nr1ol7te9i]

YOU SAVE ME ON MY EXAME I HOPE YOU HAVE GOOD TIME ALL TIME

[@emillytabara9410]

This guy it's just awesome!

[@KyleFES]

When deciding between addition rule and product rule I go by, if it is 'and' it is multiply, if it is 'or' it is addition.

[@timding6241]

For question b, does n means: the number of trial/attempt (I choose 4 numbers, so n=4?), or the number of available options (i.e., 8). How to answer this question if there are 0,1,2,3,4,5,6,7,8,9? Should n become 5 or something. I am confused here. Sometimes, it seems that n is the number available, sometimes, it is like n is the number of attempts. Thank you.

[@umarhaidar9492]

ur a legend m8

[@abhijitbhat2760]

When do u use combination and when to use simple probability methods??

[@sp0onshtd200]

What if there are more “code spaces” then avaliable digits

[@rikdebbanerjee55]

eddie sir...hello this is rikdeb from india..your lectures in this channel have reall helped to move forward with mathematics.. i cant do a particular sum on circular arrangements and it would be great if you attempt to help me out a little... the problem goes as follows.. What is the number of ways in which 6 men and 5 women can be arranged in around table so that no 2 women are side by side.. i really cant approach this sum by the way you usually teach in your lectures of fixing a chair for a unit at the very begginning..so could you in your free time (if available ) help me out or atleast give a few hints? your help will be greatly appreciated and wish you tons of luck for winning the global teacher's prize 2018 and also congrats for beings it's top 10 finalists... you really are a great sir... thank you..

[@yashwanth7209]

First, arrange the 6 men in a circle. There are (6−1)!=5! ways to arrange them since the arrangement is circular. Now, we need to place the 5 women in the spaces between the men. To ensure that no two women are seated side by side, we can place them in the alternate spaces between the men. There are 6 spaces between the men, so the number of ways to arrange the women in these spaces is 6!.Finally, we multiply these two results to get the total number of arrangements: Total arrangements=5!×6!

[@SankalpaSatyal]

❤

[@TheTanmaybishnoi]

I Wouldn’t bunk any of your classes :3

[@jingcao9296]

For part A, if there are no restrictions, why wouldn't it be 8^4 divided by 4! times 4!

[@ExtremerZocker284076]

The response is a little late 😅 But the reason is: You can't pick the same number twice. That's why you don't use 8^4, but 8!= 8×7×6×.... In the denominator, the first 4! is because you only have 4 spots to fill, so (n-k)!=4!. The second 4! is because you don't care about the order since it's a combination and not a permutation. Hope it helped :)

[@christopherpdearing]

@@ExtremerZocker284076 Another late response, but a combination that uses the same number twice or more isn't 8^4, it's a combination with repetition. There's a special formula for that. I think a combination with no restrictions should use that formula, not the one in the video. 8^4 would describe a scenario where abc is a different outcome than cba. When considering combinations, those are the same.

[@berukberhanu3369]

For condition C: why wouldn't it be 4C4 for the even numbers?

[@rafiashemonti4099]

It is wdym

[@marcusnguyen8600]

all of ur students prob are all band 6s

[@mekharajkunwarchhetri249]

For the last one. Can we do 4c1x7c3, as from 4 odd numbers we have to choose 1. Since we took 1 number from set we have to choose 3 numbers from remaining 7 numbers. And we have choose them simultaneously so multiply. I know the answer will be more than the sample space so it's wrong, but i don't know how it got into my head. Can anyone help by explaining ?

[@thiagomfdn]

You are blending subsets (at least three elements before and after the multiplication), which would mean that the events are not equally likely. Think odds and evens as two separete equaly likelly events(subsets), so the answer would be in the sum: 4c1x4c3 + 4c2x4c2 + 4c3x4c1 + 4c4. I hope this helps, if you are still looking for an answer