Solution :
The function :
be a random permutation.
f is a permutation on
, 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
Therefore, total number of permutation functions
are 10!.
Now we want the total number of permutation functioning :
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
, f(0)=0 and f(1)=1 are 8!
Now we want the probability that a random permutation
satisfies f(0) = 0 and f(1) = 1.
The number of permutation function
, i.e.
The probability that a random permutation
satisfies f(0) = 0 and f(1) = 1 is
![$\frac{8!}{10!} = \frac{8!}{10 \times 9\times 8!} =\frac{1}{10 \times 9}=\frac{1}{90}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B8%21%7D%7B10%21%7D%20%3D%20%5Cfrac%7B8%21%7D%7B10%20%5Ctimes%209%5Ctimes%208%21%7D%20%3D%5Cfrac%7B1%7D%7B10%20%5Ctimes%209%7D%3D%5Cfrac%7B1%7D%7B90%7D%24)
Therefore, the probability that a random permutation
satisfies f(0)= 0 and f(1)=1 is