I am trying very hard to understand following inclusion-exclusion problem but can't get it. It will be very helpful if someone can provide detail explanation.

f(s) is number of ways of having sum of elements equal to s in set T

g(i,j) is number of ways to get sum upto i using j distinct numbers from 1 to N.

Can anyone explain how

$f(n) = \sum_{i \ge 1} (-1)^i*g(n,i)$

This problem is actually from competitive programming if someone wants to refer to full resource and it's solution is also there.

  • $\begingroup$ I explained the solution, but I could not answer your question directly because what your interpretation of $f(n)$ is not correct. $s(n)$ in their solution is not actually counting anything. $\endgroup$ – Mike Earnest Jan 23 at 19:38

You are trying to count the number of solutions to $$ x_1+x_2+\dots+x_{n} = k $$ which satisfy $0\le x_a\le a-1$ for all $1\le a \le n$.

If we replaced this with the weaker condition $0\le x_a$, the answer would be $\binom{k+n-1}k$. Here is where inclusion-exclusion comes into play; you take all $\binom{k+n-1}k$ solutions, subtract the "bad" solutions where one of the variables satisfies $x_a\ge a$, then add back in the doubly subtracted solutions where two of the variables are too big, etc.

Suppose we want to count solutions where $x_a\ge a$ for all $a\in S$. That is, all of the variables in $S$ (and possibly others) are too big. Subtracting $a$ from each of the bad variables, this is the same as solving $$ x_0+x_1+\dots+x_{n-1}=k-\Big(\sum_{i\in S}i\Big) $$ Letting $i$ be the sum in question, the number of solutions is $\binom{k-i+n-1}{k-i}$. Furthermore, if this subset has size $i$, then it its sign in the inclusion exclusion sum will be $(-1)^i$.

When performing the inclusion-exclusion, we must sum over all subsets $S$. However, we can group together all of the subsets $S$ which have the same sum and size, as this is all that is needed to compute their contribution. This is where the $f(i,j)$ comes into play; it is the number of subsets with sum $i$ and size $j$. Therefore, the final answer is $$ \sum_{i=0}^k \binom{n-1+(k-i)}{(k-i)}\sum_{j\ge 0}(-1)^j f(i,j) $$

  • $\begingroup$ Thanks for the answer. I can't upvote it due to my bad reputation. Sorry for that. $\endgroup$ – Parth Patel Jan 24 at 16:06
  • $\begingroup$ @ParthPatel You’re welcome! $\endgroup$ – Mike Earnest Jan 24 at 18:22

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