Correct Answer - Option 2 : Only II
Here, it is given that U contains 23 different compounds i.e U= 23
S is the subset of U i.e. S= 9
U\S = 23 – 9 = 14
Consider this as a graph where U is the vertex set of graph G and S is the subset of vertices in G.
Now, it is given that S reacts with exactly 3 compounds of U. It means degree of every vertex is 3.
Sum of degree of every node of S = 9 × 3 = 27
U\S contains even number of compounds.
Statement I: Each compound in U\S reacts with an odd number of compounds.
Here, if each compound in U\S reacts with odd number of compounds, the sum of degree of all the nodes in U\S will be even and sum of degree of all nodes in graph G will be odd as sum of degree of all nodes in S is odd. This is not possible using handshaking lemma. So, Statement I is false.
Statement II: At least one compound in U\S reacts with an odd number of compounds.
To satisfy the handshaking lemma for graph, sum of degree of all nodes U\S must be odd. To make this happen, we require atleast one node of U\S with an odd degree. We can assign it in such a way that sum of degree of all nodes in U\S will be odd. So, it is true.
Statement III: Each compound in U\S reacts with an even number of compounds.
If each compound in U\S reacts with an even number of compounds, sum of degree of all nodes in U\S will become even and sum of degree of all nodes in graph will become odd as sum of degree of all nodes in S is odd. And an odd number added with an even number of results in an odd number. To satisfy the handshaking lemma, sum of degree of all nodes in the graph must be even. So, it is false.
