20:16 with the (n-3)! part was completely lost on me. Could anyone shine some light on that part? I totally didn't follow how all that is the same as (n-1) choose 2.
Wait, complete graph without loop(s) but they're definitely graph with all the vertices connected, so the number of edge is combination(v,2). But why you draw that K4 only with 5 edges, not 6?
I dont know if you still want this, but he basically made the connection with (n-1) choose 2 which would be the equation that he wrote. (n-1) choose 2 is (n-1)!/(n-3)!*2 which is same as (n-1)(n-2)(n-3)!/(n-3)!*2. cancelling out the (n-3)! and you get what he was going with before he made the comparison to (n-1) choose 2.
Question about compliments Can I say that taking the compliment of a graph is just getting rid of the edges that are there, but adding the ones that aren't? You probably said it, but I think I got confused... Sorry
TheTrevTutor i just saw your video about complete graphs and you took the compliment of a complete bipartite graph and it cleared things up. Thank you!
@@JP-xm3qf But any edge must connect TWO vertices so even if there are three edges each of them must connect TWO vertices. So (4 choose 2) is basically "how many pairs of vertices one can choose" since each pair will make ONE edge.
Because in a complete graph, any vertex is connected to all other vertices. So it has n-1 edges incident on it. Thus if you remove a vertex, you have to remove n-1 edges as well.
![[Discrete Mathematics] Vertex Degree and Regular Graphs](http://i.ytimg.com/vi/k7ThfZIT_ac/mqdefault.jpg)
At 4:35 - I think you meant to say lets remove the vertex C, and not "lets remove the edge C".
Thank you sir.. You can explain better than my College lecturer.. Very Helpful !!
That is to be expected! Nothing surprising considering most of these days lecturers in college.😀
These are better than the videos my professor posts. Thank you so much
Thank you for posting these videos. They are helping me tremendously.
You are a legend
at 12:28 why are we looping it from the bottom but not looping it from the top just like the K4
It doesn't matter right?
It's a bunch of jargon!
10:00 how does K1 have an edge? K1 is an isolated vertex where the degree is 0 which means no edge. But you said differently. Thanks
20:16 with the (n-3)! part was completely lost on me. Could anyone shine some light on that part? I totally didn't follow how all that is the same as (n-1) choose 2.
Wait, complete graph without loop(s) but they're definitely graph with all the vertices connected, so the number of edge is combination(v,2). But why you draw that K4 only with 5 edges, not 6?
Damn these lectures are so ''edgy''.
20:34 I don't understand the (n-3)! part...can anyone help? Thank you!
I dont know if you still want this, but he basically made the connection with (n-1) choose 2 which would be the equation that he wrote. (n-1) choose 2 is (n-1)!/(n-3)!*2 which is same as (n-1)(n-2)(n-3)!/(n-3)!*2. cancelling out the (n-3)! and you get what he was going with before he made the comparison to (n-1) choose 2.
21:46
22:00
Question about compliments
Can I say that taking the compliment of a graph is just getting rid of the edges that are there, but adding the ones that aren't? You probably said it, but I think I got confused... Sorry
Yeah, that would be one way of looking at it.
TheTrevTutor i just saw your video about complete graphs and you took the compliment of a complete bipartite graph and it cleared things up. Thank you!
How many sub-graph does Kn? 🤔 Please
15:21 because we have 4 vertices and we chose 2??????
because an edge is formed between 2 vertices so total number of edges is 4 choose 2
But this is a complete graph, so all the vertex are connected directly, so the vertices are connected with 3 edges
@@JP-xm3qf But any edge must connect TWO vertices so even if there are three edges each of them must connect TWO vertices. So (4 choose 2) is basically "how many pairs of vertices one can choose" since each pair will make ONE edge.
6:47 Wait that's illegal!!
in 19:15 why the edge of {v} is (n-1). I do not understand
Because in a complete graph, any vertex is connected to all other vertices. So it has n-1 edges incident on it. Thus if you remove a vertex, you have to remove n-1 edges as well.
20:16 ohhhh magic
I've completely missed what he did there. Anyone able to help? Like with the (n-3)! bit...I'm totally lost...
Living on the edge be like
eyvvah 205