I want to complete the following proof: So continuing where the author left off, I did the following: $\begin{array} {cc} & \sum\limits_{i=1}^{n-1} P(A_i) – \sum\limits_{1\le i < j \le n-1}P(A_i \cap A_j) + \dots + (-1)^n P(A_1 \cap A_2 \cap \dots \cap A_{n-1}) + P(A_n) – P(\bigcup\limits_{i=1}^{n-1} (A_i \cap A_n))\\ &= \sum\limits_{i=1}^{n-1} P(A_i) – \sum\limits_{1\le […]

How many integer solution to the inequality $$y_1+y_2+y_3+y_4\leq184$$with $y_1>0,\,0<y_2\leq10,\,0\leq y_3\leq17 $ and $0\leq y_4< 19$ MY try:For two variable I could try to solve by graphical method but for this, I don’t know.Thank you

Find the number of soultion to this equation: $$x_1+x_2+x_3+x_4+x_5+x_6= 20$$ $$1 <= x_1, x_2, \ldots, x_6 \leq 4 $$ I know that at the start we need to define: $y_1=x_1-1$ $y_2=x_2-1$ $y_3=x_3-1$ $y_4=x_4-1$ $y_5=x_5-1$ $y_6=x_6-1$ So we end up with this: $$y_1+y_2+y_3+y_4+y_5+y_5=14$$ and now I need to use the “Inclusion–exclusion principle” but how?

I’m an amateur games designer, and am working on a mechanic which involves rolling on a table with numbers running from 2-12 – the full range of possibilities from adding together two six-sided dice. Calculating the probability of any given result on 2d6 (or any other number of dice, when all of them are summed) […]

I made a sequence of terms using the inclusion exclusion principle and odd primes and was wondering whether it converged to $\frac{n}{2}$, which was my original hope. It looks like this: $$\lfloor\frac{n}{6}\rfloor+\lfloor\frac{n}{10}\rfloor+…+\lfloor\frac{n}{2p}\rfloor- (\lfloor\frac{n}{30}\rfloor+… \lfloor\frac{n}{2p_i p_{i+1}}\rfloor+…) +\lfloor\frac{n}{210}\rfloor+\lfloor\frac{n}{2p_i p_{i+1} p_{i+2}}\rfloor+…-…+…-\frac{n}{2*\prod_{k=2}^{\pi(n)}p_k}$$ I hope the pattern is clear here. In your answer, please discuss why or why not the […]

This question is motivated by the elaboration of the question Combinatorial Proof of Inclusion-Exclusion Principle (IEP). Let’s consider the following two aspects: 1.) IEP transforms at least information to exact information Applying the Inclusion-Exclusion Principle to $n$ sets $A_1,\dots,A_n$ gives: \begin{align*} \left|\bigcup_{j=1}^{n}A_j\right|=\sum_{j=1}^{n}\left|A_j\right| -\sum_{1\leq i \leq j \leq n}\left|A_i\cap A_j\right|\pm\cdots+(-1)^{n-1}\left|\bigcap_{j=1}^{n}A_j\right|\tag{1} \end{align*} We observe that the LHS […]

Does anyone know a reference for the solution of the generalized derangement problem via Laguerre polynomials? The Wikipedia article here says that this is an application of inclusion-exclusion, but I don’t see how. This formula was used by joriki in a nice MSE answer here. The article below solves the generalized derangement problem with inclusion-exclusion, […]

The answer seems to be yes. Please let me know if I have made a mistake in my argument or if there is an opportunity to make the argument simpler, clearer, or more standard. Let $a \ge 1, n \ge 2$ be integers. Let $\frac{n}{4}\#$ be the primorial for $\left\lfloor\frac{n}{4}\right\rfloor$ Let $\gcd(a,b)$ be the greatest […]

I recently read Doron Zeilberger’s paper on Inclusion-Exclusion Principle. Let’s say there are $n$ properties which are numbered $1,\cdots,n$. And let $A$ be a set of elements which has some of these properties. Then the Inclusion-Exclusion Principle states that the number of elements with no properties at all is \begin{equation*} \sum_{I\subset \{1,\cdots,n\}} (-1)^{|I|}\cdot |A_I| \end{equation*} […]

How to prove following Identity? $$\sum_{k=0}^n (-1)^k {n-k \choose k} m^k (m+1)^{n-2k} = \frac {m^{n+1}-1}{m-1}, m \ge 2$$ This seems very hard to me. Any idea about how to prove it combinatorialy? [P.S: for $k > \lfloor {\frac n 2}\rfloor$ L.H.S evaluates to $0$. ]

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