# Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done isanswer:53

Latika Leekha
9 years ago
Hello student,
Number of derrangements of 6 = 6! (1 – 1/1! + 1/2! - 1/3! + 1/4! – 1/5! + 1/6!)
= 360 – 120 + 30 – 6 + 1
= 265
Out of these derrangements, there are five ways in which card numbered 1 is going wrong.
So, when it is going in envelope numbered 2 is 265/5 = 53 ways.
Gautham Krishna
26 Points
7 years ago
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
ankit singh
4 years ago
Hello friend,
Number of derrangements of 6 = 6! (1 – 1/1! + 1/2! - 1/3! + 1/4! – 1/5! + 1/6!)
= 360 – 120 + 30 – 6 + 1
= 265
Out of these derrangements, there are five ways in which card numbered 1 is going wrong.
So, when it is going in envelope numbered 2 is 265/5 = 53 ways.