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Not getting highlighted part how ceil N/K will be greater or equal to r?

closed with the note: Got the point
in Combinatory by Loyal (6.5k points)
closed by | 23 views
lets take an example:

There is K boxes (k=5). And N objects (N=6). pigeonhole-principle says that at least one Box contain r objects (r=2).Then

                      ceilOf(N/K) ≥ r

                     ceilOf(5/4)  ≥ r

                                   2  ≥ r   // r=2 which is true

                   ceilOf(N/K) can be greater than r Here we take only least amount of objects

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