Изменения

|statement=Пусть <tex> A = \bigcup_bigcup \limits_{i=1}^{n}A_i </tex> , тогда по формуле включения-исключения: <center> <tex> | A | = \sum_sum \limits_{I=(i_1,i_2, \ldots ,i_k) \subset \{ 1,2, \ldots ,n \} } (-1)^{k+1} \Big| \bigcap_bigcap \limits_{ j \in I } A_j \Big| </tex> </center>

<tex> A = \bigcup_bigcup \limits_{i=1}^{n}A_i = \Bigg( \underbrace {\bigcup_bigcup \limits_{i=1}^{n-1}A_i}_{B} \Bigg) \cup A_n </tex>

<tex> | B | = \sum_sum \limits_{I \subset \{ 1,2, \ldots ,n-1 \} } (-1)^{|I|+1} \Big| \bigcap_bigcap \limits_{ j \in I } A_j \Big| </tex>

<tex> \Big| B \bigcap A_n \Big| = \Bigg| \Bigg( \bigcup_bigcup \limits_{i=1}^{n-1}A_i \Bigg) \bigcap A_n \Bigg|= \Bigg| \bigcup_bigcup \limits_{i=1}^{n-1} \bigg( A_i \bigcap A_n \bigg) \Bigg| = </tex>

<tex> = \sum_sum \limits_{I \subset \{ 1,2, \ldots ,n-1 \} } (-1)^{|I|+1} \bigg| \bigcap_bigcap \limits_{ j \in I } \Big( A_j \bigcap A_n \Big) \bigg| = \sum_sum \limits_{I \subset \{ 1,2, \ldots ,n-1 \} } (-1)^{|I|+1} \Big| \bigcap_bigcap \limits_{ j\in I \cup \{ n \} } A_j \Big| </tex>

<tex> | A | = | A_n| + \Bigg( \sum_sum \limits_{I \subset \{ 1,2, \ldots ,n-1 \} } (-1)^{|I|+1} \Big| \bigcap_bigcap \limits_{ j \in I } A_j \Big| \Bigg) - - \Bigg( \sum_sum \limits_{I \subset \{ 1,2, \ldots ,n-1 \} } (-1)^{|I|+1} \Big| \bigcap_bigcap \limits_{ j\in I \cup \{ n \} } A_j \Big| \Bigg) = \sum_sum \limits_{I \subset \{ 1,2, \ldots ,n \} } (-1)^{|I|+1} \Big| \bigcap_bigcap \limits_{ j \in I } A_j \Big| </tex>