حل تشریحی سوال شماره 251 ریاضیات گسسته ریاضی فیزیک

روش اول:

(0/25)                  $3^5-\left(3\times 2^5-3\right)=150$

                             (0/25)    (0/25)    (0/25)

روش دوم:

(0/25)                  $\begin{array}{c}{A}_1=f:A\to B|f\left(a_1\right)\ne b_1,1\le i\le 5\Rightarrow \left|A_1\right|=2^5\\ A_2=f:A\to B|f\left(a_1\right)\ne b_2,1\le i\le 5\Rightarrow \left|A_2\right|=2^5\\ A_3=f:A\to B|f\left(a_1\right)\ne b_3,1\le i\le 5\Rightarrow \left|A_3\right|=2^5\end{array}\}$

$|A_1\cap A_2|=|A_1\cap A_2|=1(0/25),A_1\cap A_2\cap A_3|=0$

$\left|\overline{A_1\cup A_2\cup A_3}\right|=\left|A_1\cup A_2\cup A_3\right|\left(0/25\right)=3^5-\left(3\times 2^5-3\times 1\right)=150\left(0/25\right)$