Prove the Dirichlet box principle If n objects are put into

Prove the Dirichlet box principle: If n objects are put into m boxes, some box must contain [n/m] objects, and some box must contain 6 [n/m].

Solution


We shall use the method of induction. According to which we will assume the contradiction and prove it wrong.
Let us suppose that total \"n\" number of objects are to be put in \"m\" number of boxes and n>m.

Let us assume that there is no box with at least n/m objects

In this case, each and every box will have less than n/m objects

Hence
Number of objects in each box < nm

Total number of objects< number of boxes x nm

Total number of objects < m × nm

Total number of objects < n
But given that number of objects are strictly equal to n.
Which is a contradiction to our assumption.
Hence there exists at least one box having at least nm objects

This proves the generalized form of Dirichlet principle.

 Prove the Dirichlet box principle: If n objects are put into m boxes, some box must contain [n/m] objects, and some box must contain 6 [n/m].Solution We shall

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