Answer:
In the same way, to have a particular pair \((a_i,a_j)\) fixed, we can choose any permutation of the remaining \(n - 2\) elements; there are \((n - 2)!\) such choices and thus \[P(A_i \cap A_j) = \frac{(n - 2)!}{n!} = \frac 1{n(n - 1)}\ .\] The number of terms of this form in the right side of Equation [eq 3.5] is \[{n \choose 2} = \frac{n(n - 1)}{2!}\ .\] Hence, the second term of Equation [eq 3.5] is \[-\frac{n(n - 1)}{2!} \cdot \frac 1{n(n - 1)} = -\frac 1{2!}\ .\] Similarly, for any specifi…
Step-by-step explanation:
