Respuesta :
Solution :
The function : [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] be a random permutation.
f is a permutation on [tex]$Z_{10}$[/tex] , i.e. f is permutation 10.
Now we know that the total number of distinct permutation on to symbolize 10!.
Each of these 10! permutation to a permutation function [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex]
Therefore, total number of permutation functions [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] are 10!.
Now we want the total number of permutation functioning :
[tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] such that f(0) = 0 and f(1)= 1
Now we notice that when f(0)=0 and f(1)=1, then two symbol '0' and '1 are fixed under permutation f.
So essentially when f(0) = 0 and f(1) = 1, f becomes permutation on 8 symbol.
Total number of permutation functioning [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] , f(0)=0 and f(1)=1 are 8!
Now we want the probability that a random permutation [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] satisfies f(0) = 0 and f(1) = 1.
The number of permutation function [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] , i.e.
The probability that a random permutation [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] satisfies f(0) = 0 and f(1) = 1 is
[tex]$\frac{8!}{10!} = \frac{8!}{10 \times 9\times 8!} =\frac{1}{10 \times 9}=\frac{1}{90}$[/tex]
Therefore, the probability that a random permutation [tex]$f: Z_{10} \rightarrow Z_{10}$[/tex] satisfies f(0)= 0 and f(1)=1 is [tex]$\frac{1}{90}$[/tex]