Why didn't we have such cool teachers... You are so energetic and made me want to come to your class ... That's some very cool teaching ... Loved it . (And thanks for solving my doubt lol)
The quote “A Problem Well Stated is Half Solved” is usually attributed to Charles Kettering, who was head of research at General Motors from 1920 to 1947.
cheesayn hi Chessany The problem is that In this case we only have 1 item ( is a Girl) rather than 3 items in your cmt!!! Then think about it, how many way to place 1 item (a Girl) in 3 areas (3 sits) :D Further explanation for 3! Most easily way to think about 3! is use example. How many ways can you arrange 3 items (3 people) into 3 area (let say 3 different roles such as President, Vice President and Secretary)? The answer could be 3 ways for President, 2 ways for Vice and 1 ways for Secretary ==> 3! Same example, still 3 areas (3 different roles like above), but now only have 1 item (1 people), so how many different roles that the people can be? (Assuming that 1 people only can get 1 role only)!! ==> 3 roles (3 ways) only !! Hope this can help you to understand and get the answer for your question !!!
@mehawaarahoon the reason why it's 3 is because we are calculating the no. Of possible arrangements. If we straight up put 1x0x0 it meant, we are opting for a particular arrangement. Over here the girl has 3 seats to choose from and that's why we multiply by 3. Further, it's only a single person being put in place, hence we don't use factorial because after placing her in a seat, the problem is solved.
How is 5!^2 hard I don't understand. Calculations are not that difficult in these to do mentally and be called brilliant for that. He teaches excellent though I admit! Calculations anyone can do without calculator if you've reached to the Combinatorics part!!
Interesting approach to the last question. I would have found the options for 4 couples round a table (96) and then inserted the boy into one of the 4 spots between couples, and the girl into one of the treee remaining spots between couples (96x4x3). Same answer of course, but different thought process.
Another method:- 3!............( for 4 couples) (2!)^4........(for arrangement of these couples) Now there are 4 gaps between all the 4 couples hence selecting 2 gaps and distributing 2 people among them, so 4P2 Hence the answer=> 4P2 ×(2!)^4×3!=1152
The x3 in the last question can be considered in nPr as 3P1, meaning 3!/(3-1)!, therefore 3!/2! which simplifies to 3. if we make a mistake and write 3! instead of x3, it would mean 3P3 which essentially is filling 3 seats with 3 people, when we only need to fill 1 person.
part (d) Can I consider that the bays and the girls are two couples , the first couple has 5 boys and the second couple has 5 girls , And therefore we have 2 couples each couple has 5 persons ways = (2-1)! * 5 ! * 5 !
Part e one pair of boy - girl cannot be seated together , the other 4 boys and 4 girls have no restrictions or they couples too and must be seated together?
For the example where one couple may not sit together we may also use complementary counting. Hence seat the four couples and the two singles without restrictions and remove the cases in which the singles are together: (5!*2^4)-(4!*2^5)
Because it doesn't make a difference. As it's not a row but a circle, when you place 1 person and view it from any given perspective, it'll always be the same, hence we only multiply it by 1. This is done, in order to fix the following positions for other sitting arrangements.
The only things making these seats distinguishable from one another are the objects occupying them. That means that in a roundtable, two empty chairs aren't different from one another but two occupied chairs are. So like if I sat down in a roundtable with one red chair and one blue chair and I sat down on the blue chair, it would be considered the same as sitting on the red chair because at the end of the day, sitting on either of those chairs would leave you with one empty chair in front of you. It doesn't matter what the surroundings are or how different the chairs are, what matters is what's on the seats. Two seats are only different if there are two different objects on each of them. That's why sitting on any of those 5 seats wouldn't have made a difference in circular permutation, because it all would've been the same anyways, you'd have one empty chair to your left, one empty chair to your right, one empty chair to your front-left, and one empty chair to your front-right. Basically in circular permutation, if you have the same things around you, it's considered the same position. that's how I think it works at least
Can someone explain me why 120 or 5P5 times 5P5 divided by 10 and all times 2 is the answer for boys and girls sit alternatly which is 4! times 5! in the video?
for part (d), they are able to rotate 1 seat (as an entire group) 9 times, without breaking up each boy/girl group... like a clock. thus wouldn't you multiply what you have by 9, as well??
The solution should be the same as when solving for alternating boys and girls because there aren't any couples left so we treat them as individual units
error u cant place the girl in 3 ways, u can only place her in 2 if u place her two seats to the right of the intial boy, then u cant place all the couples down
Nope, I think what is given in the video is correct. Think about it this way: each spot represents a place where the boy, girl, or a couple can sit. At this stage, we are not thinking of the couple as 2 people, but rather as a single entity. He considers the order of the couple internally in (2!)^4 later on.
I can't imagine that I am watching such high-quality explanations for free!
I can. It's called youtube. What did you just watch..
You are probably the best teacher ever! The explanation, delivery, simplicity, everything is phenomenal and highly engaging! Kudos to you Sir!
thank !! you!! so!! much!! for making and putting together this blessed playlist because permutations and combinations are very confusing !!
Why didn't we have such cool teachers... You are so energetic and made me want to come to your class ... That's some very cool teaching ... Loved it . (And thanks for solving my doubt lol)
The quote “A Problem Well Stated is Half Solved” is usually attributed to Charles Kettering, who was head of research at General Motors from 1920 to 1947.
At the end: "Why is it not 3! ?"
Thanks for asking that, Dude :)
cheesayn hi Chessany
The problem is that In this case we only have 1 item ( is a Girl) rather than 3 items in your cmt!!!
Then think about it,
how many way to place 1 item (a Girl) in 3 areas (3 sits) :D
Further explanation for 3!
Most easily way to think about 3! is use example. How many ways can you arrange 3 items (3 people) into 3 area (let say 3 different roles such as President, Vice President and Secretary)?
The answer could be 3 ways for President, 2 ways for Vice and 1 ways for Secretary ==> 3!
Same example, still 3 areas (3 different roles like above), but now only have 1 item (1 people), so how many different roles that the people can be? (Assuming that 1 people only can get 1 role only)!! ==> 3 roles (3 ways) only !!
Hope this can help you to understand and get the answer for your question !!!
@@huytangquoc5682 Thanks dude! Your explanation and example really helped me understand that guy's question 😃
@mehawaarahoon the reason why it's 3 is because we are calculating the no. Of possible arrangements. If we straight up put 1x0x0 it meant, we are opting for a particular arrangement. Over here the girl has 3 seats to choose from and that's why we multiply by 3. Further, it's only a single person being put in place, hence we don't use factorial because after placing her in a seat, the problem is solved.
you said it well Sir, a problem explained well is solved half, I always struggle more to understand the question, than solving it.
1:51 Kids, be thankful that you have very brilliant teacher. He just calculated 5!^2 MENTALLY!
p.s thank you Mr. Woo!
Knowing 5! is 120 easy to figure out.
12x12 is something kids can do in primary school.
120x120 is trivially harder.
How is 5!^2 hard I don't understand. Calculations are not that difficult in these to do mentally and be called brilliant for that. He teaches excellent though I admit! Calculations anyone can do without calculator if you've reached to the Combinatorics part!!
Crystal clear in one shot.
Interesting approach to the last question.
I would have found the options for 4 couples round a table (96) and then inserted the boy into one of the 4 spots between couples, and the girl into one of the treee remaining spots between couples (96x4x3).
Same answer of course, but different thought process.
Another method:-
3!............( for 4 couples)
(2!)^4........(for arrangement of these couples)
Now there are 4 gaps between all the 4 couples hence selecting 2 gaps and distributing 2 people among them, so
4P2
Hence the answer=> 4P2 ×(2!)^4×3!=1152
I solved it this way too
Seriously, your teaching skills are amazing. Thank you so much sir. Well explained in simple ways.
Best explanation on internet 👍 thanks a lot ☺️
The x3 in the last question can be considered in nPr as 3P1, meaning 3!/(3-1)!, therefore 3!/2! which simplifies to 3.
if we make a mistake and write 3! instead of x3, it would mean 3P3 which essentially is filling 3 seats with 3 people, when we only need to fill 1 person.
Thank you so much for explaining something so complex in a simple and funny way😍♥️♥️
You are the best teacher on TH-cam
Oh god, I hate perms and combs. Can't wait to revise the hard questions with your videos!
@@misterwootube Thank u, sir
My Hsc is in three days and I'm surviving off these
how did you go?
part (d) Can I consider that the bays and the girls are two couples , the first couple has 5 boys and the second couple has 5 girls ,
And therefore we have 2 couples each couple has 5 persons
ways = (2-1)! * 5 ! * 5 !
Excellent explanation. Simple and clear.
Thank you very much!
Part e one pair of boy - girl cannot be seated together , the other 4 boys and 4 girls have no restrictions or they couples too and must be seated together?
Can anyone plz explain me this....in case (d) the set of chairs can any different right? If yes why are we not considering them?
For the example where one couple may not sit together we may also use complementary counting. Hence seat the four couples and the two singles without restrictions and remove the cases in which the singles are together: (5!*2^4)-(4!*2^5)
the why he teaches the topic i think no one else can't do better than him
how about arranging the objects 1, 1, 1, 2, 2, 2, 1, 1, 1 around a table?
You're really good!!
sir i dont understand that you fixed on person if example i have 5 boys then for fixed that boy i have 5 ways why u did not multiply by 5
Because it doesn't make a difference. As it's not a row but a circle, when you place 1 person and view it from any given perspective, it'll always be the same, hence we only multiply it by 1. This is done, in order to fix the following positions for other sitting arrangements.
The only things making these seats distinguishable from one another are the objects occupying them. That means that in a roundtable, two empty chairs aren't different from one another but two occupied chairs are. So like if I sat down in a roundtable with one red chair and one blue chair and I sat down on the blue chair, it would be considered the same as sitting on the red chair because at the end of the day, sitting on either of those chairs would leave you with one empty chair in front of you. It doesn't matter what the surroundings are or how different the chairs are, what matters is what's on the seats. Two seats are only different if there are two different objects on each of them. That's why sitting on any of those 5 seats wouldn't have made a difference in circular permutation, because it all would've been the same anyways, you'd have one empty chair to your left, one empty chair to your right, one empty chair to your front-left, and one empty chair to your front-right. Basically in circular permutation, if you have the same things around you, it's considered the same position.
that's how I think it works at least
GOD BLESS YOU
part d) shouldn't the whole be multiplied by 2! as we can fix the boys on the other 5 seats as well??
no coz it's the same. It's a round table or a circle so doing that is simply just rotating the circle which is hence the same. Hope that helped :)
2! Is used because the couples can interchange the seats with themselves.
Can someone explain me why 120 or 5P5 times 5P5 divided by 10 and all times 2 is the answer for boys and girls sit alternatly which is 4! times 5! in the video?
I wonder how we would have solved if there were 2 breakups?
for part (d), they are able to rotate 1 seat (as an entire group) 9 times, without breaking up each boy/girl group... like a clock. thus wouldn't you multiply what you have by 9, as well??
only 4 years late, but no, as if you rotate them like a clock they are still in the same "order"
How many ways would we have if all couples broke up?
The solution should be the same as when solving for alternating boys and girls because there aren't any couples left so we treat them as individual units
But, what if I want to know how many ways we have if everyone don’t want to seat alongside his/her former couple.
Thank you man!
🔥 🔥
Lovely!
Good teacher.
love u sirrrr
nice nice nice
error u cant place the girl in 3 ways, u can only place her in 2
if u place her two seats to the right of the intial boy, then u cant place all the couples down
Nope, I think what is given in the video is correct. Think about it this way: each spot represents a place where the boy, girl, or a couple can sit. At this stage, we are not thinking of the couple as 2 people, but rather as a single entity. He considers the order of the couple internally in (2!)^4 later on.
@@stavanjain7894 thanks! I had the same doubt!
Understood it now.
Thank u, sir
👍
Any one who is teacher here?