Following the result that the number of vertices having odd degree in the degree sequence must be even to be a valid graphic sequence.But here we have 5 vertices having degree 3 i.e. odd degree ,hence the given sequence is not a graphic sequence.Hence S1 is false.
Given that we have not a cycle of length 3.So minimum length cycle that we can have = 4.
Let there are r such cycles (or regions) .
every region is bounded by 4 edges at least
So in r regions we have e >= 4r/2 [ We have divided by 2 to avoid double counting of edges]
⇒ e >= 2r
Now we are given that the graph is planar and connected,so Euler's theorem holds.So by Euler's theorem,
r = e - n + 2
Substituting in the earlier result we have ,
e >= 2(e - n + 2)
⇒ e <= 2n - 4
So we have n = 10 and e = 16 which on substitution on the above inequality gives :
16 <= 16 which is true.
Hence S2 is true.
Hence , option B) is true