Mathematics PROBLEM HELP!

G is a connected plane graph with n vertices each of degree a, where n and a are positive
integers. Every face of G has degree either a or a − 1.

Find the number of faces of each degree in terms of n and a.

Any thoughts/ideas? Where are you stuck?

i know the sum of degrees of all faces = 2m which means m = na/2

eulers formulae leads to f = 2 + na/2 - n (total faces)

im not sure where to go from here to find each degree faces

OK, so you have one equation for the number of faces - a good start.

Since we're interested in the number of faces of degree a, and a-1 (the only options); if, we let them be x, y respectively, we have f=x+y, which you can replace in the above.

Can you get another equation relating x,y to n and a, by considering the degrees of the faces?

so i actually got to the f = x+ y part also, it was the question you asked me i dont know why i am struggling haha

OK, perhaps I was being too cryptic - I'll rephrase it.

You have x faces of degree a, and y faces of degree a-1. Can you relate that information to the total number of edges?

na = a + a-1
so a = 1/n-2
so a-1 = (1/n-2) -1


?

Not quite.

Total number of edges, counted twice, is na, yes.

But if you have x faces of degree a, how many edges is that? Similarly y.

x,y faces of degree a is

a + x -2 edges
a + y -2 edges

Not sure what your thinking is there.

The degree of a face is the number of edges it has.

So, a face of degree "a" will have "a" edges.
2 faces of degree "a" will have 2a edges.

And so, x faces of degree a will have ax edges.

Similarly y with a-1 edges.

Can you contiue from there to get the second equation in x,y.

so i got na/2 = ax + y(a-1)

Close.

Each edge belongs to two faces, so when you count the number of edges belonging to each face and sum over all the faces you end up counting each edge twice.

Hence, na = ax + y(a-1)

Note: For the final answers, they've asked for them to be in terms of n,a, so best not to have "f" there.

Close.

Each edge belongs to two faces, so when you count the number of edges belonging to each face and sum over all the faces you end up counting each edge twice.

Hence, na = ax + y(a-1)

Note: For the final answers, they've asked for them to be in terms of n,a, so best not to have "f" there.

so,
y = 2a + na^2/2 - 2na
x = 2 + na/2 -n -2a -na^2/2 + 2na


part b say does the formulae hold for all possible values of n and a

REVISED ANSWER, how does part b work?