I would like to ask a question! Hope you could answer me~ 4 different red balls and 5 different blue balls are placed in a row in a container with 9 partitions. Find the number of ways of arranging them if a) the red balls and blue balls are placed alternately b) no red balls are placed next to each other
(a) 4! x 5! , the arrangement of the ball can only be b1, r2 , b3, r4 , b5 , r6 ,b7 ,r8 , b9 so we are just calculating the different possibilities of arrangement of the four red colour balls in their red balls position and the same for the blue one . (b) ( 4! x 5! ) x 2 , the first arrangement of the four red balls can be done by putting them at the 1st , 3rd , 5th , 7th partitions , and then the second type of arrangement of the red balls is by putting them at the 2nd , 4th , 6th , and 8th , and if I'm not mistaken , R1R3R5R7 are just the same as R3R5R7R9 , but I'm not sure about it though , you better check the answers to confirm this
I am struggling with one question, can you help: There are 10 seats in a row in a cinema theater. 6 people are shown to this row and two of them must sit together next to each other. In how many different ways can they be seated? According to me it should be 2 seats taken by the couple leaves 8 seats for the 4 people. the first person has 8 options, the second has 7 , then 6 and then 5. The other two have 10 and 9. I am confused. Please reply
So firstly you must group the couple together (the 2 of them) -> Because they now take up "1 seat", you can permutate them 9 times out of the 10 original seats. After which, they can both exchange positions so you permutate the 2 of them. Next, you have the remaining 4 of them and with the remaining 8 seats, so permutate the 4 of them out of the 8 seats (8 choose 4 then multiply 4!). Add everything and you should be good to go.
@@levupeducation So if my understanding is correct, 9!x2!= 725760, the second part is 8C4x4!= 1680. Now if we add up both values we get 727440. But the answer in the book is 30240. I still don't get it.
@@jcutey Since 2 seats have been taken up by the couple and no matter where they sit they have to be together, you can treat the 2 people as 1 person. Now with this logic, only 1 seat has been taken up. Now there are 9 seats left, 4 individual people and 1 couple (taken as 1 person) to be arranged among these 9 seats. Therefore its 9P5 = 15120. Since the 2 people can exchange positions, you permutate the 2 of them (2!). Therefore the answer is 9P5*2!=30240 :)
Hi for the insertion method is it not possible to insert them at the extreme right and left? Since they will not be standing beside each other. Would it be 8C2 instead of 6C2?
Yup I totally agree with this oneee , this is how I learnt at school. And in total there’s 9 people , 7 can sit with each other (that’s why we permutate 7!) 2 cannot sit with each other (that’s why we do 8C2 x 2!
God loves you and he wants to save everyone, but in order for him to do that, you need to repent and be baptized. Also share his gospel with everyone you come in to contact with and keep his commandments 🙏🏾😘
This is by far the best permutation video I've watched.
ah my nemesis lol
thanks for this haha
this was a really helpful video; thank you so much! :)
Love the lecture ❤ first time understanding online
thanks so much .....but i would like to get some qns about the topic for practice
I have a question, in (18:14) so basically you multiplied (7!) x (6C2) x (2!) and that would be the answer? Thanks for the great content!!!
Yes, correct.
@@levupeducation Hi should it not be 8C2 x 2! cuz you did not include slotting it before A and after G
@@user-rv1gv9pl8u I'm assuming it had to be within the extreme boundaries and not outside them
How can we solve a given data that that two and three different types of repetition e.g BABALAWO?
Good explanation 👏 👍
I would like to ask a question! Hope you could answer me~
4 different red balls and 5 different blue balls are placed in a row in a container with 9 partitions. Find the number of ways of arranging them if
a) the red balls and blue balls are placed alternately
b) no red balls are placed next to each other
(a) 4! x 5! , the arrangement of the ball can only be b1, r2 , b3, r4 , b5 , r6 ,b7 ,r8 , b9 so we are just calculating the different possibilities of arrangement of the four red colour balls in their red balls position and the same for the blue one . (b) ( 4! x 5! ) x 2 , the first arrangement of the four red balls can be done by putting them at the 1st , 3rd , 5th , 7th partitions , and then the second type of arrangement of the red balls is by putting them at the 2nd , 4th , 6th , and 8th , and if I'm not mistaken , R1R3R5R7 are just the same as R3R5R7R9 , but I'm not sure about it though , you better check the answers to confirm this
@@alvinkhoo2193 Thx a lot!!
Thank you☆
great video!in 8:23 we also could use the method 10P5 too right?
just to make it sure>
Yes, absolutely!
Tahnkyouuuu💕
I am struggling with one question, can you help: There are 10 seats in a row in a cinema theater. 6 people are shown to this row and two of them must sit together next to each other. In how many different ways can they be seated? According to me it should be 2 seats taken by the couple leaves 8 seats for the 4 people. the first person has 8 options, the second has 7 , then 6 and then 5. The other two have 10 and 9. I am confused. Please reply
So firstly you must group the couple together (the 2 of them) -> Because they now take up "1 seat", you can permutate them 9 times out of the 10 original seats. After which, they can both exchange positions so you permutate the 2 of them.
Next, you have the remaining 4 of them and with the remaining 8 seats, so permutate the 4 of them out of the 8 seats (8 choose 4 then multiply 4!).
Add everything and you should be good to go.
@@levupeducation So if my understanding is correct, 9!x2!= 725760, the second part is 8C4x4!= 1680. Now if we add up both values we get 727440. But the answer in the book is 30240. I still don't get it.
Can you send the question to my email? alevellessons@gmail.com
@@jcutey Since 2 seats have been taken up by the couple and no matter where they sit they have to be together, you can treat the 2 people as 1 person. Now with this logic, only 1 seat has been taken up. Now there are 9 seats left, 4 individual people and 1 couple (taken as 1 person) to be arranged among these 9 seats. Therefore its 9P5 = 15120. Since the 2 people can exchange positions, you permutate the 2 of them (2!). Therefore the answer is 9P5*2!=30240 :)
It becomes easy if you forget the couple and consider it as one person, hope this helps :)
Hi for the insertion method is it not possible to insert them at the extreme right and left? Since they will not be standing beside each other. Would it be 8C2 instead of 6C2?
Yup I totally agree with this oneee , this is how I learnt at school. And in total there’s 9 people , 7 can sit with each other (that’s why we permutate 7!) 2 cannot sit with each other (that’s why we do 8C2 x 2!
Hey, you never talked about if items in a row a being permutated and a pair a placed next to each other how you work around it
Basically attachments with restrictions
Arrangements ^
Hey ! ! Can you tell me what do you mean by r that you specified
The number of items chosen from the total number or items available.
For identical objects, what if I only
Take
3 out of 6, then how to solve?
3P6 divide by (the number of identical object A)! x (the number of identical objects B)! x … x (the number of identical object N)!
are u singaporean ?? istg u sound just like my singaporean friend!
Yes I am! But i don't think i am your friend hahaha ;)
I don't understand the circle one
did not understand a single thing from permutations !
Hope this video clarified some doubts !
God loves you and he wants to save everyone, but in order for him to do that, you need to repent and be baptized. Also share his gospel with everyone you come in to contact with and keep his commandments 🙏🏾😘
who fucking asked
@@botdemonitisation2527Jesus loves you
your accent makes it impossible to understand the video