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There are $20$ students in a class, where every two students have the same grandfather. Prove that there is a grandfather which has at least $14$ grandchildren from the class.

If we represent $20$ students as one set of vertices, and grandfathers as another set, we don't necessarily have a bipartite graph. Instead, we have a disconnected graph that contains $10$ components (trees with three vertices). But, we have to prove that there is a grandfather which has at least $14$ grandchildren.

What should be the topology of this graph?

How to prove this?

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    In this day and age, you should make sure to spell out clearly that every student has *at most* two grandfathers :) – 2017-01-22
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    Let's assume that no student is a grandfather. Then the graph you mention with students on one side and grandfathers on the other side satisfies all properties of a bipartite graph (even if it is disconnected, which it is not since every two students share one grandfather). – 2017-01-22

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You don't need graph theory at all. Suppose each grandfather has at most 13 grandchildren. Let $c_1$ be one of grandchildren of $f_1$. There are at least 7 students who are not grandchildren of $f_1$. Let $c_2$ be one of them. And they all have common grandfather $f_2$ with $c_1$ (otherwise $c_1$ would have at least three grandfathers). Now there are at least 7 students who are grandchildren of $f_1$ and not grandchildren of $f_2$. Let $c_3$ be one of them. Then $c_2$ and $c_3$ have the common grandfather $f_3$. Also $f_3$ should be grandfather of all 7 students who are not grandchildren of $f_1$ (otherwise $c_2$ would have at least three grandfathers). On the other hand $f_3$ should be grandfather of all 7 students who are not grandchildren of $f_2$ (otherwise $c_3$ would have at least 3 grandfathers). We've got at least $7 + 7 = 14$ grandchildren of $f_3$ that is a contradiction, which proves the statement.