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Let $M=\{i:x\in A_i\}$, with $|M|=m$. Then $1_{A_S}(x)=1$ exactly when $S\subseteq M$, and the sum becomesThis is $1$ when $m=t$ and $0$ otherwise. Summing over $x\in X$ yieldsTaking the complement of the $t=0$ case gives the inclusion-exclusion principle

If $p_1,\dots ,p_r$ are the distinct prime divisors of $N$, inclusion-exclusion removes their multiples from $\{1,\dots ,N\}$ and givesFor the stated squarefree $n=q_1\cdots q_k$ and $\gcd (x,n)=1$, Fermat's theorem gives $x^{q_j-1}\equiv 1(\mathrm{mod}{q}_j)$. Since $q_j-1\mid n-1$, every $q_j$ divides $x^{n-1}-1$; their product does too. Thus $n$ is a Carmichael number.

The inclusion-exclusion principle states that for finite sets $A_1,\dots ,A_k$,For each prime $p\mid n$, exclude pairs whose two coordinates are divisible by $p$. For a set of distinct prime divisors $p_1,\dots ,p_j$, exactly $n^2/(p_1\cdots p_j{)}^2$ pairs have both coordinates divisible by their product. Inclusion-exclusion therefore gives

For events $A_1,\dots ,A_n$, the probabilistic inclusion-exclusion principle isTo prove it, fix an outcome lying in exactly $r$ events. Its total coefficient on the right isby the binomial theorem; outcomes in no event contribute zero. Taking expectations of this pointwise indicator identity proves the formula. Truncating after the pair terms gives the Bonferroni inequalities

For the final bound, let $X={\sum }_i{1}_{A_i}$ and $S=\mathbb{E}X={\sum }_i\mathbb{P}(A_i)$. The intersection assumption givesThe Cauchy-Schwarz inequality applied to $X{1}_{X>0}$ yields $S^2\le \mathbb{E}[X^2]\mathbb{P}(X>0)\le S+n-1$. Thereforeso the required universal constant may be taken as $\boxed{c=2}$. This is a second moment method estimate.

The inclusion-exclusion principle says that the size of a finite union is the alternating sum of the sizes of all nonempty intersections of its constituent sets.

For each prime number $p\mid n$, let $A_p$ consist of the tuples for which $p$ divides every $x_i$. Exactly $(n/p{)}^r$ tuples lie in $A_p$, and for distinct primes $p_1,\dots ,p_j\mid n$, exactly $(n/(p_1\cdots p_j){)}^r$ lie in their intersection. A tuple has greatest common divisor greater than one with $n$ exactly when it belongs to some $A_p$. Inclusion-exclusion therefore gives the Jordan totient function