Let $n\ge 3$ be a natural number. How many possibilities do we have to find 3 odd natural numbers $k_1,k_2,k_3>0$ such that $k_1+k_2+k_3\leq n$?
A simple question about counting possibilities
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combinatorics
1 Answers
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you want $2a+1+2b+1+2c+1\leq n$ so $2a+2b+2c\leq n-3$ so $a+b+c\leq \lfloor \frac{n-3}{2}\rfloor$.
The number of solutions in non-negative integers to $a+b+c\leq x$ is clearly equal to the number of solutions in non-negative integers to $a+b+c+d=x$, by stars and bars this is $\binom{x+3}{3}$.
So you want $\binom{\lfloor (n-3)/2 \rfloor +3}{3}$.