Each 2 represents a "flip" switch meaning there is a choice whether to have the order BG or GB. In this case. we have 5 separate couples (switches), so there are a total of 2^5 ways the switches can be flipped or not.
@@renielsulit6985 no it doesn't matter whether a boy sits first or a girl as whosoever sits on 1st chair will find 4 people from the similar sex and 5 people of the other sex sitting at the table
@@theunique140 See in couple can sit in (2!) Ways and there r 5 couples. when two Or more things happen simultaneously, we multiply them instead of adding them, hence 2! Has to be multiplied 5 times which means (2!) raised to power 5 Hope u understood 😊
Ahr dam, i thought i found what i was looking for, fixed binary necklace aka how many ways to arrange boys and girls around the table without differing which boy or girl. (So that altering between boy and girl would have only 1 solution). It is so hard to find. Anyway, just wanted to say that your lessons are brilliant. It is a wonder that you do not have more views, especially with that many subscribers.
Why is the first example only 1× 9! ? I thought the first person could be one out of those 10 people so it is 10 ×... and he has 10 places to choose so 10×10×9! 😅
Because there is a fundamental difference in the circular case - namely, that each seat is identical. If the 10 people were sat around a table with no rotational symmetry, then there would indeed be 10! permutations as each seat is different. Another way to explain this is that in the non-symmetrical case, there are 10! = 10 x 9 x 8 x ... x 1 permutations. You can think about this as there being 10 places for the first person to sit, 9 places for the second, etc. Notice that by the time we've placed 9 people, we've actually got all of the possible permutations, since there are no options remaining for the last person - there is one seat left, and he has to sit in it! Similarly, when we place the first person in our circular case it actually doesn't matter where he sits, since every seat is identical. Once the first person has sat down, though, the remaining 9 seats are no longer identical - we can uniquely define them by their relationship to the first seat. Hence there are 1 x 9 x 8 x 7 x ... x 1 = 9! permutations. Yet another way of thinking about this is given by Eddie in the video. If you initially treat the problem as you would any other permutation problem, then you identify that there are 10! permutations. However, for each of the unique permutations (the ones we actually care about), there are 9 other permutations which are generated by progressively rotating the table 1/10 of the way around. You can see from the diagram that these 9 additional permutations are not unique - we can obtain them by simply rotating our viewpoint from the initial permutation. From the point of view of any of the seats, they would have no idea which of the 10 permutations they were in - remember, there are no external "landmarks" to act as a reference, so the only way to know which permutation you are in is to look at the people in the surrounding seats. This means that we have over-counted by a factor of 10, so the total number of unique permutations becomes 10!/10 = 9! Hopefully this helps!
Why is it not 10! ? I don't understand why it needs to be just 9! If you're over counting because order doesn't matter, then why does order not matter only for 10 permutations? What is so special about those specific 10 permutations that doesn't occur in all the 9! rest of them
Pineapple29 the difference between 10! and 9! is not 10 permutations. When you calculate for 10! you are taking into account the permutations which are not different due to relative change between the items/individauls but those achieved through rotation.
Why is it raised to 5, rather than multiplying by 5?
Each 2 represents a "flip" switch meaning there is a choice whether to have the order BG or GB.
In this case. we have 5 separate couples (switches), so there are a total of 2^5 ways the switches can be flipped or not.
does it matter if boy sits first or girl sit first in b/g alternate? shouldnt it be x2!?
@@renielsulit6985 no it doesn't matter whether a boy sits first or a girl as whosoever sits on 1st chair will find 4 people from the similar sex and 5 people of the other sex sitting at the table
@@NishantKumar-jq2gh I also didn't understand why 2! raise to 5?
@@theunique140 See in couple can sit in (2!) Ways and there r 5 couples. when two Or more things happen simultaneously, we multiply them instead of adding them, hence 2! Has to be multiplied 5 times which means (2!) raised to power 5
Hope u understood 😊
so glad i did the last question on my own and it turned out to be correct
Thank you for this
It has helped me a lot
Thank you very much!
I don’t comment on videos much but thanks this helps so much
Thank you!!!
What would be the answer in items a) and b) if the seats were identified with different letters?
Ahr dam, i thought i found what i was looking for, fixed binary necklace aka how many ways to arrange boys and girls around the table without differing which boy or girl.
(So that altering between boy and girl would have only 1 solution). It is so hard to find.
Anyway, just wanted to say that your lessons are brilliant. It is a wonder that you do not have more views, especially with that many subscribers.
Why is 5P5 times 5P5 divided by 10 and doubled or times two the answer for arrangements of boys and girls sit alternatly or same as 4! times 5!
Really nice and smooth explanations
I miss you
Why is the first example only 1× 9! ? I thought the first person could be one out of those 10 people so it is 10 ×... and he has 10 places to choose so 10×10×9! 😅
Because there is a fundamental difference in the circular case - namely, that each seat is identical. If the 10 people were sat around a table with no rotational symmetry, then there would indeed be 10! permutations as each seat is different.
Another way to explain this is that in the non-symmetrical case, there are 10! = 10 x 9 x 8 x ... x 1 permutations. You can think about this as there being 10 places for the first person to sit, 9 places for the second, etc. Notice that by the time we've placed 9 people, we've actually got all of the possible permutations, since there are no options remaining for the last person - there is one seat left, and he has to sit in it! Similarly, when we place the first person in our circular case it actually doesn't matter where he sits, since every seat is identical. Once the first person has sat down, though, the remaining 9 seats are no longer identical - we can uniquely define them by their relationship to the first seat. Hence there are 1 x 9 x 8 x 7 x ... x 1 = 9! permutations.
Yet another way of thinking about this is given by Eddie in the video. If you initially treat the problem as you would any other permutation problem, then you identify that there are 10! permutations. However, for each of the unique permutations (the ones we actually care about), there are 9 other permutations which are generated by progressively rotating the table 1/10 of the way around. You can see from the diagram that these 9 additional permutations are not unique - we can obtain them by simply rotating our viewpoint from the initial permutation. From the point of view of any of the seats, they would have no idea which of the 10 permutations they were in - remember, there are no external "landmarks" to act as a reference, so the only way to know which permutation you are in is to look at the people in the surrounding seats. This means that we have over-counted by a factor of 10, so the total number of unique permutations becomes 10!/10 = 9!
Hopefully this helps!
we fixed the first person on one seat.
Because to the first person all chairs aresame
hiya!
Teacher, where r u from?
Great content. Easy to understand.
why you stop the video in between
Why aren't we counting the number of various different couples which can be formed
The couple here means that they are a pair. Would you seat with your girlfriend or some other girls?
Why is it not 10! ? I don't understand why it needs to be just 9!
If you're over counting because order doesn't matter, then why does order not matter only for 10 permutations? What is so special about those specific 10 permutations that doesn't occur in all the 9! rest of them
Pineapple29 the difference between 10! and 9! is not 10 permutations. When you calculate for 10! you are taking into account the permutations which are not different due to relative change between the items/individauls but those achieved through rotation.
not great explanation
then watch previous of his circular videos...... his teachings are linked