2. In how many ways can we choose two different integers between -100 and 100 inclusive, so that their sum is greater than their product?
SolUtion
Let the two integers be $m$ and $n$ and we let $mm n
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This can be rearranged into the useful inequality below because each of the unknowns is now written only once.
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(m-1)(n-1)m \geq m n$, so all these cases work. There are $100 \times 100=10000$ of these.
If $m=00=m n$, so this always works. There are 100 of these cases.
If $m=1n=m n$, so this always works. There are 99 of these cases.
If $1<m<n$ we have $m+n<n+n=2 n \leq m n$, so this never works.
In total, there are $10000+100+99=10199$ ways.