1 , 2 , 3 , 4 , 5
This 5 number can purmuted in 5! ways=120 ways
Now, 1 cannot be placed in 1st, 3 cannot be placed in 3rd, 5 cannot be placed in 5 th.
Now, think all are in their position.
So, remaining 2 element can be placed as 2! ways.=2...................i
Now, think only 1 and 3 in their position.
Other elements can place in 3! ways.......................ii
But we have to remove te case i from case ii
So, here 3!-2! ways.
Now, similar case for 3,5 in their position and 1,5 in their position
So, $3\times (3!-2!)$ways..............iii
At last we have to think only 1 in it's position.
so, here we have to remove case i and case ii as duplicate case
Say, for 1 is in place
From case ii we are getting similar case when (1,3) is in position, and when (1,5) is in position and also case i , when all 1,3,5 all are in position.
$4!-(2\times (3!-2!)-2!)-2!$
Similar case if only 3 is in position and only 5 in position
i.e.$(4!-(2\times (3!-2!)-2!)-2!)\times3.$............iv
So, remaining case are 120-(2!+$(3!-2!)\times3$+$(4!-(2\times (3!-2!)-2!)-2!)\times3$.)=120-2-12-18=88