+2 votes
180 views
Consider $2n$ committees, each having at least $2n$ persons, formed from a group of $4n$ persons. Prove that there exists at least one person who belongs to at least $n$ committees.
asked in Others
edited | 180 views

## 1 Answer

+4 votes
Best answer

This problem is modeled as a bipartite graph.

1. $2n$ vertices represent $2n$ committees.
2. $4n$ vertices represent $4n$ people. We have,

\begin{align*} &\Rightarrow \sum d_i = 2E \\ &\Rightarrow {\color{red}{\sum d_{x_i}}} + {\color{blue}{\sum d_{y_i}}}= 2E \\ &\Rightarrow {\color{red}{\sum d_{x_i}}} + {\color{blue}{\sum d_{y_i}}} = 2.{\color{red}{\sum d_{x_i}}} \\ &\Rightarrow {\color{blue}{\sum d_{y_i}}} = {\color{red}{\sum d_{x_i}}} \\ &\Rightarrow {\color{blue}{\sum_{\;}^{4n} d_{y_i}}} = \left ( 4n^2 \right )_{\text{min}} \\ &\Rightarrow 4n.\left ( d_{\text{people}} \right )_{\text{average}} = \left ( 4n^2 \right )_{\text{min}}\\ &\Rightarrow \left ( d_{\text{people}} \right )_{\text{average}} = n_{\text{min}} \\ \end{align*}

\begin{align*} &\Rightarrow \exists y \in {\text{\{People\}}} \text{ such that } d_y \geq n \\ \end{align*}

answered by Veteran (58k points)
selected

+11 votes
2 answers
1
0 votes
1 answer
2
+3 votes
4 answers
3
0 votes
0 answers
5
+3 votes
1 answer
7