Any one has any idea how to prove it? "if we put $n$ items into $n-1$ boxes then there will be a box with $2$ items" I have tried like Base case: for $n=1\implies 1$ item, $1$ box Hypothesis: for $n=k \implies k$ items , $k-1$ boxes Inductive step: $n=k+1\implies 1+2+3+...+k+(k+1)=\frac{(k+1)(k+1+1)}{2}$ ....but I think I am wrong.
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First: the statement is false as written: all the items could be in one of the boxes. You want to say "with at least $2$ items".
To establish the inductive step suppose you have $n+1$ items and $n$ boxes. If there is a box with just one item in it, throw away that box and the item in it and use the inductive hypothesis.
Figure out a similar argument if there is an empty box.
This is not the kind of (induction) problem you attack with formulas.
Ethan Bolker
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