2 votes 2 votes 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. Combinatory isi2016-pcb combinatory descriptive + – Devasish Ghosh asked Mar 5, 2017 • retagged Nov 9, 2019 by Lakshman Bhaiya Devasish Ghosh 589 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 4 votes 4 votes This problem is modeled as a bipartite graph. $2n$ vertices represent $2n$ committees. $4n$ vertices represent $4n$ people. We have, $$\begin{align*} &\Rightarrow \sum d_i = 2E \\ &\Rightarrow {\color{red}{\sum d_{x_i}}} + {\color{blue}{\sum d_{y_i}}}= 2E \\ &\Rightarrow {\color{red}{\sum d_{x_i}}} + {\color{blue}{\sum d_{y_i}}} = 2.{\color{red}{\sum d_{x_i}}} \\ &\Rightarrow {\color{blue}{\sum d_{y_i}}} = {\color{red}{\sum d_{x_i}}} \\ &\Rightarrow {\color{blue}{\sum_{\;}^{4n} d_{y_i}}} = \left ( 4n^2 \right )_{\text{min}} \\ &\Rightarrow 4n.\left ( d_{\text{people}} \right )_{\text{average}} = \left ( 4n^2 \right )_{\text{min}}\\ &\Rightarrow \left ( d_{\text{people}} \right )_{\text{average}} = n_{\text{min}} \\ \end{align*}$$ $$\begin{align*} &\Rightarrow \exists y \in {\text{\{People\}}} \text{ such that } d_y \geq n \\ \end{align*}$$ dd answered Mar 5, 2017 • selected Mar 5, 2017 by Devasish Ghosh dd comment Share Follow See all 0 reply Please log in or register to add a comment.