Thank you for your videos, really helpful with getting back into Discrete Math after a 15-year break. I have found an easier way to think about the circular tables problem, at least for me. So we have 16 people to arrange, and let us assume we have 16 total chairs we need to fill in. We can find the number of arrangements by a simple 16 * 15 * 14 * ... * 2 * 1 which is 16! and all we need to do is to divide that number by 10 for the possible number of the around the table rotations (non-new arrangements) of the first table and by 6 for the possible number of the around the table rotations (non-new arrangements) of the second table. So the TOTAL would be equal to 16! / 10 * 6 which is basically a bit more simplified version of your answer.
I should have started watching this series back when my semester began, it makes things a lot clearer... probably mainly because it isn't like the class I am taking, where everything is crammed into 1-a-week 3 hour classes that take place at night. ~_~ FML
It doesn't matter which table you choose first because C(16,10)=C(16,6). This works for any two numbers N and M where they sum to the top number in the binomial. C(M+N,N)=C(M+N,M). If CW and CCW directions are not important, you need to divide your answer by 4.
Thank you for this, I assume there is a proof I could look up somewhere describing the formula you laid out here. I was confused on why those were equal when they clearly aren't the same numbers. Working through a discrete math class now and only in week 2. Thank you again.
Hi Trev, love your videos! One thing I always get confused about is when to multiply vs. add. In your last problem when I tried it on my own before you showed the answer, I added the 3 expressions when they should have been multiplied. For your last problem in the "Permutation Practice" video, we added up all of the cases. Just curious if there is an intuitive way to think about when to multiply vs when to add? You are the best.
@03:05 RRRRUUUUU (or as u said EEEENNNNN ) since in each step we should move either one step to the right or one to the up, then it means RRRRUUUUU must be arranged in allternative, i.e URURURURU, ISN'T IT!? WHICH MEANS 5!X4!
First of all, thanks. Your videos are super helpful, as always. I have one question about the circular table exercise. My initial approach was slightly different. I am getting a different answer, and I am trying to figure out why. My train of thought is: I start sitting people at table A (10 seats), one by one. Then, I proceed to table B (6 seats). So, let's start. For the first table, I can sit people in 1 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 ways. Continuing with the second table, that is, 1 x 5 x 4 x 3 x 2 x 1. After simplifying, that is (15!/6!) x 5! which gives an answer smaller than yours. I understand that I am missing cases but I don't know which ones. Any idea what I am missing here? Thanks! 🙂
hi, I have question 2:37 if I change Q3 a bit => how many words can be made that contain ConsVowelCons? Is the answer 3!*C(4,4)*4!*C(4,2)*2! ? My thought was place the three vowels first, then chose four Cons fill the interval blank, at last fill the remain Cons in the interval. Am I correct or not?
@05:40 How many ways can we sit 16 people around two circular tables, one holds 10 and another holds 6 seats!? I'm afraid that your answer is not correct, becuase we can start we the smaller table too, isn't it? starting with big table or small table= ur answer+....
I will make it two cases: Case 1: Sit the people first in the big table= Ur answer. Case 2: Sit the people first in the small table= Ur answer because ((16 choose 10) = (16 choose 6)) All possible ways!? Case+Case, isn't it!?
I got it. So once that we used (16 choose 10) it means we used all 16 people, for the first table, isn't it? Now!? what about the second table? You used only (6 choose 6) did you use all 16 people in the arrangement of the second table, no, isn't it?
In the two circular tables question, if the question says "how many ways can we sit 16 people around two circular tables" wouldn't this mean that you would have to account for people changing tables (ie. in one arrangement person x could be on the 10 seater and an another arrangement person x could be on the 6 seater)? Or am I missing where this has already been taken into account?
@@Trevtutor thank you. However, why, then, is there the need for 6C6? I mean, 6C6 = 1 and 16C6 accounts for all the possible ways. So I don’t see the need for 6C6? Or am I missing something?
In the question "(-1, 0, 6) -> (3, 2, 8) where x -> x+1, y->y+1 or z ->z+1 in each step" dont you need to change out "or" with "xor"? If not we dont specify only 1 movement each step? With or you could go both in x and y at the same time, right?... Or am I misunderstanding something?
"If not we dont specify only 1 movement each step". That should be clear and go without saying after introducing walks with only two options in the previous video/question.
I am a little confused for question 1 part 3. "How many cons, vowel, cons words can be made." Isn't this question similar to the problem in your last video where you were inserting different letters into TALLAHASSEE? When I did this problem before you explained it, I did (3!) for the order of the vowels, multiplied by, (5 choose 3) since there are only 5 positions where there can be cons | vowel | cons and 3 potential vowels, and then multiplied by (6!) for the order of the consonants. Yet that is completely wrong. Could you please explain this further? Thanks
+Muyiwa Olaniyan It's not the same arrangement as TALLAHASSEEE. in TALLAHASSEE I'm rearranging the whole word. In the question in this video, I'm only taking 3 letters from the list to make a word. ConsVowelCons (or CVC), are words like cat, gij, gat, gej, etc.
In the how many permutations question. Shouldnt it be 1 x 3 x 5 ... because 4 letters is already used. So 9 - 4 used = 5. Or if j can be used again it should be 1 x 3 x 6 ... because 9 - 3 = 6. Please explain
I believe you could have also done 16P10/10( how many ways to arrange 10 people out of 16 at the first table) and then 6P6/6(how many ways to arrange 6 people out of the number of people left(16-10=6)) and then multiplied it like you did. Is that an ok way to think of the problem?
If you get the same answer, then I suppose it's fine. It might be more difficult to adapt to other problems though, where you might want 4 girls and 6 guys at table 1, then 4 girls and 2 guys at table 2.
I am having a little difficulty understanding the circular table logic. Why are we dividing with the number of possible rotations? Assume we have a triangular table so 3 people can sit. Total number of arrangements is 3! but with the circular logic is it only 3??? Below is the total arrangements: 1 1 2 2 3 3 2 3 3 2 1 3 3 1 1 2 2 1
You don't have 6 different arrangements there. If we take 1 2 3 and rotate the table so that 2 is at the top, we get 2 3 1 Because the table only sits one person at each edge, we have to count rotations as being the same seating arrangement. So in your case, 3!/3 = 2 different arrangements.
What if there were less people than 16 in the last example? Say 14 people in two tables of 10 and 6? Anyways, great video! Thanks for the great quality content!
I think it's because the circular table with x amount of seats implies that for every arrangement we have around that table, we will also end up with x amount of "copies" of that arrangement that, because it's circular, we treat as all the same arrangement (because no matter how you turn a circular table, what could be at seats A, B, C, D, etc. could always move a seat down and it'd be the same arrangement) -- unless we divide by x to eliminate all x "copies".
Hey everyone I hope you are doing alright just I wanna say that GOD loved the world so much he sent his only begotten son Jesus to die a brutal death for us so that we can have eternal life and we can all accept this amazing gift this by simply trusting in Jesus turning away from your sins and forming a relationship with heavenly father.
Thank you for your videos, really helpful with getting back into Discrete Math after a 15-year break. I have found an easier way to think about the circular tables problem, at least for me. So we have 16 people to arrange, and let us assume we have 16 total chairs we need to fill in. We can find the number of arrangements by a simple 16 * 15 * 14 * ... * 2 * 1 which is 16! and all we need to do is to divide that number by 10 for the possible number of the around the table rotations (non-new arrangements) of the first table and by 6 for the possible number of the around the table rotations (non-new arrangements) of the second table. So the TOTAL would be equal to 16! / 10 * 6 which is basically a bit more simplified version of your answer.
You are a great teacher.
I should have started watching this series back when my semester began, it makes things a lot clearer... probably mainly because it isn't like the class I am taking, where everything is crammed into 1-a-week 3 hour classes that take place at night. ~_~ FML
It doesn't matter which table you choose first because C(16,10)=C(16,6). This works for any two numbers N and M where they sum to the top number in the binomial. C(M+N,N)=C(M+N,M).
If CW and CCW directions are not important, you need to divide your answer by 4.
Thank you for this, I assume there is a proof I could look up somewhere describing the formula you laid out here. I was confused on why those were equal when they clearly aren't the same numbers. Working through a discrete math class now and only in week 2. Thank you again.
in those graphical questions, we can use manhattan distance to find the path
Hi Trev, love your videos! One thing I always get confused about is when to multiply vs. add. In your last problem when I tried it on my own before you showed the answer, I added the 3 expressions when they should have been multiplied. For your last problem in the "Permutation Practice" video, we added up all of the cases. Just curious if there is an intuitive way to think about when to multiply vs when to add? You are the best.
We add when the question says "a *or* b." We multiply when there's an "and" present.
@03:05 RRRRUUUUU (or as u said EEEENNNNN ) since in each step we should move either one step to the right or one to the up, then it means RRRRUUUUU must be arranged in allternative, i.e URURURURU, ISN'T IT!?
WHICH MEANS 5!X4!
no? You can do either, so why would it have to alternate?
I see,
I just understood it wrong.
How do you know when to multiply or add? For instance, in the last question I thought of adding the two combinations.
For every possible combination for 1st table all combinations on the second table are valid
@@1Eagler You add when you can t do the 2 or more operations at the same time. You multiply when you can do the operations at the same time.
First of all, thanks. Your videos are super helpful, as always. I have one question about the circular table exercise. My initial approach was slightly different. I am getting a different answer, and I am trying to figure out why. My train of thought is: I start sitting people at table A (10 seats), one by one. Then, I proceed to table B (6 seats). So, let's start. For the first table, I can sit people in 1 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 ways. Continuing with the second table, that is, 1 x 5 x 4 x 3 x 2 x 1. After simplifying, that is (15!/6!) x 5! which gives an answer smaller than yours. I understand that I am missing cases but I don't know which ones. Any idea what I am missing here? Thanks! 🙂
For two tables question, it can also be considered as (16 * 15 * 14 * ... * 6)/10 * (6!/6)
hi, I have question 2:37 if I change Q3 a bit => how many words can be made that contain ConsVowelCons?
Is the answer 3!*C(4,4)*4!*C(4,2)*2! ?
My thought was place the three vowels first, then chose four Cons fill the interval blank, at last fill the remain Cons in the interval.
Am I correct or not?
wait I was wrong, is it (6*3*5)*6!*C(7,1)? I'm lost...
ha, you had answered it in Counting Practice course 😁
@05:40 How many ways can we sit 16 people around two circular tables, one holds 10 and another holds 6 seats!?
I'm afraid that your answer is not correct, becuase we can start we the smaller table too, isn't it?
starting with big table or small table= ur answer+....
(16 choose 10) = (16 choose 6). It doesn't matter.
Yeah!
Still! we can do both
either this or that (ur answer+ur answer)
So ur answer X 2 !?
I will make it two cases:
Case 1: Sit the people first in the big table= Ur answer.
Case 2: Sit the people first in the small table= Ur answer because ((16 choose 10) = (16 choose 6))
All possible ways!? Case+Case, isn't it!?
No. You're not doing two cases, you're simply re-ordering which table is seated first, which makes no difference to the overall question.
I got it.
So once that we used (16 choose 10) it means we used all 16 people, for the first table, isn't it?
Now!? what about the second table? You used only (6 choose 6) did you use all 16 people in the arrangement of the second table, no, isn't it?
Trev can you upload a counting #of passwords video from easy to difficult
In the two circular tables question, if the question says "how many ways can we sit 16 people around two circular tables" wouldn't this mean that you would have to account for people changing tables (ie. in one arrangement person x could be on the 10 seater and an another arrangement person x could be on the 6 seater)? Or am I missing where this has already been taken into account?
It's accounted for with the choose operation. (16 choose 10) picks all possible sets of 10 at one table, and 6 at another.
@@Trevtutor thank you. However, why, then, is there the need for 6C6?
I mean, 6C6 = 1 and 16C6 accounts for all the possible ways. So I don’t see the need for 6C6? Or am I missing something?
i can't understand in circular table's question why we rotate everyone one position it need divide the number of total people
In the question "(-1, 0, 6) -> (3, 2, 8) where x -> x+1, y->y+1 or z ->z+1 in each step" dont you need to change out "or" with "xor"? If not we dont specify only 1 movement each step? With or you could go both in x and y at the same time, right?... Or am I misunderstanding something?
"If not we dont specify only 1 movement each step". That should be clear and go without saying after introducing walks with only two options in the previous video/question.
i didnt understand how we approach the questions
Ben de ahahahah :D
I am a little confused for question 1 part 3. "How many cons, vowel, cons words can be made." Isn't this question similar to the problem in your last video where you were inserting different letters into TALLAHASSEE? When I did this problem before you explained it, I did (3!) for the order of the vowels, multiplied by, (5 choose 3) since there are only 5 positions where there can be cons | vowel | cons and 3 potential vowels, and then multiplied by (6!) for the order of the consonants. Yet that is completely wrong. Could you please explain this further? Thanks
+Muyiwa Olaniyan It's not the same arrangement as TALLAHASSEEE. in TALLAHASSEE I'm rearranging the whole word. In the question in this video, I'm only taking 3 letters from the list to make a word. ConsVowelCons (or CVC), are words like cat, gij, gat, gej, etc.
+TheTrevTutor Ahh got it. Thanks!
In the how many permutations question. Shouldnt it be 1 x 3 x 5 ... because 4 letters is already used. So 9 - 4 used = 5. Or if j can be used again it should be 1 x 3 x 6 ... because 9 - 3 = 6. Please explain
I believe you could have also done 16P10/10( how many ways to arrange 10 people out of 16 at the first table) and then 6P6/6(how many ways to arrange 6 people out of the number of people left(16-10=6)) and then multiplied it like you did. Is that an ok way to think of the problem?
If you get the same answer, then I suppose it's fine. It might be more difficult to adapt to other problems though, where you might want 4 girls and 6 guys at table 1, then 4 girls and 2 guys at table 2.
Hi I don't understand how you know when to use the division rule?
I am having a little difficulty understanding the circular table logic. Why are we dividing with the number of possible rotations?
Assume we have a triangular table so 3 people can sit. Total number of arrangements is 3! but with the circular logic is it only 3???
Below is the total arrangements:
1 1 2 2 3 3
2 3 3 2 1 3 3 1 1 2 2 1
You don't have 6 different arrangements there. If we take
1
2 3
and rotate the table so that 2 is at the top, we get
2
3 1
Because the table only sits one person at each edge, we have to count rotations as being the same seating arrangement. So in your case, 3!/3 = 2 different arrangements.
What if there were less people than 16 in the last example? Say 14 people in two tables of 10 and 6? Anyways, great video! Thanks for the great quality content!
Why do you have to rotate the people around circular table?
I think it's because the circular table with x amount of seats implies that for every arrangement we have around that table, we will also end up with x amount of "copies" of that arrangement that, because it's circular, we treat as all the same arrangement (because no matter how you turn a circular table, what could be at seats A, B, C, D, etc. could always move a seat down and it'd be the same arrangement) -- unless we divide by x to eliminate all x "copies".
Who else here is FU-Berlin info ersti?
Can't get it
i am 4 yrs too late but there should be 5 vowels in english = a,e, i, o, u. For some reason he only considers a,e,i as vowels
It’s only from the letters at the top
Hey everyone I hope you are doing alright just I wanna say that
GOD loved the world so much he sent his only begotten
son Jesus to die a brutal death for us so that we can have eternal life and we can all accept this amazing gift this by simply trusting in Jesus turning away from your sins and forming a relationship with heavenly father.
i didn't understand how we approach the questions