Our interest is in finding |T| where
T = {f ∈ S : f(i) is not equal to = i, for i = 2, 3, 4, 5, 6}
We classify the functions
in T into types depending upon whether f(2) = 1 or f(2) is not = 1. So, let
T1 = {f ∈ T : f(2) = 1}                                (1)
and T2 = {f ∈ T : f(2) is not = 1}                 (2)
A function in T1 interchanges 1 and 2 and maps the remaining symbols
3, 4, 5, 6 to themselves bijectively, without any fixed points. So it is
like a derangement of these four symbols. Hence
|T1| = D4 = 9                                           (3)
as calculated above. Now consider a function f in T2. This corresponds
to a bijection from the set {2, 3, 4, 5, 6} to the set {1, 3, 4, 5, 6} in which
f(2) is not = 1 and f(i) is not = i for i = 3, 4, 5, 6. If we relabel the element 1
in the codomain as 2 (because we thought that 2 not to go into 1), then f is nothing but a derangement of the five
symbols 2, 3, 4, 5 and 6. Therefore
|T2| = D5 = 60 − 20 + 5 − 1 = 44              (4)
Adding (3) and (4) we get
|T| = |T1| + |T2| = 9 + 44 = 53