I found this problem in my textbook. The formulation is as follows:
Let $A_1, \cdots , A_n - $ be events and quantities $S_0, S_1, \cdots, S_n$ are defined as follows: $S_0 = 1$
$$S_r = \sum \limits_{J_r}P(A_{k_1}\cap\cdots\cap A_{k_r}); 1 \le r \le n$$
where the summation is over all unordered $r$-tuples $J_r=\{k_1, \cdots, k_r\}$ of disctinct elements of the set $\{1, \cdots, n\}.$
Let $B_m - $ be an event that means that exactly m events from $A_1, \cdots, A_n$ will occur at the same time. Prove that
$$P(B_m)=\sum\limits_{r=m}^{n}(-1)^{r-m}{r\choose m} S_r$$
Ok, we can rewrite $B_m$ as union of all events which mean "exactly m events from $A_1, \cdots, A_n$ will occur at the same time":
$$B_m = \bigcup\limits_{J_m}(A_{k_1}\cap\cdots\cap A_{k_m})$$
After that I tried to use inclusion–exclusion principle for $B_m$ $$P(\bigcup\limits_{J_m}(A_{k_1}\cap\cdots\cap A_{k_m})) = \sum\limits_{J_m} P(A_{k_1}\cap\cdots\cap A_{k_m}) - \sum_{\alpha>\beta \in J_m} P((A_{\alpha_1}\cap\cdots\cap A_{\alpha_m}) \cap (A_{\beta_1}\cap\cdots\cap A_{\beta_m})) + \cdots$$
First addend $$\sum\limits_{J_m} P(A_{k_1}\cap\cdots\cap A_{k_m}) = {m\choose m} S_{m} = S_{m}$$ is obvious.
But I don't know how to cope with the second and etc. So, how to prove, that $$\sum_{\alpha_r>\cdots>\alpha_1\in J_m} P((A_{\alpha_{11}}\cap\cdots\cap A_{\alpha_{1m}})\cap \cdots \cap (A_{\alpha_{r1}}\cap\cdots\cap A_{\alpha_{rm}}))= (-1)^{r-m}{r\choose m} S_r$$?
