6 cartons is an odd number 1 2 3 4 5 6 and feed her to be placed in the envelope so that each envelope contain exactly one card and 2 card is placed in the and the Bear in the same number and moreover the card number one is always placed in the handle of number to then the number of ways it can be done is

							
 
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