The numbers are 1 2 3 4 5.
Now we need to make sure that 1 ,3,5 are not in their orignal position.
Lets us find the total possiblity that they are in orignal position.
a) 1 is in orignal position =4!=24

b) 3 is in orignal position=4!=24

c)5 is in orignal position=4!=24

d)1&3 are in orignal positon=3!=6

e)3&5 are in orignal position=3!=6

f)1&5 are in orignal position=3!=6

g)1&3&5 are in orignal position=2!=2
total cases in which either 1,3,5 are in correct position=24*3-6*3+2

The numbers are 1 2 3 4 5.

Now we need to make sure that 1 ,3,5 are not in their orignal position.

Lets us find the total possiblity that they are in orignal position.

a) 1 is in orignal position =4!=24

b) 3 is in orignal position=4!=24

c)5 is in orignal position=4!=24

d)1&3 are in orignal positon=3!=6

e)3&5 are in orignal position=3!=6

f)1&5 are in orignal position=3!=6

g)1&3&5 are in orignal position=2!=2

total cases in which either 1,3,5 are in correct position=24*3-6*3+2

=56.

So answer =120-56=64