da what????????

3/8

Use Derangement formula

Let  denote number of ways in which all n objects can be put in wrong positions.

If n=1 then  is obviously =0 since for 1 object there is only one position and therefore it can be placed in that position only.

If n=2 then  is again obviously =1 one way will be correct and the other way will be wrong.

If n=3 then =Total no of ways to arrange 3 objects - (no. of ways in which 1object goes in correct position and 2 in wrong positions) - (no. of ways in which 2 objects go in correct position and 1 in wrong position) - (no. of ways in which 3 objects go in correct position)

=3! - C(3,1).  - C(3,2). -1 = 6-3-0-1=2

Similarly, = 4! - C(4,1) - C(4,2) - C(4,3) - 1 = 24 - 8 - 6 -1 = 9

Total no. of ways to arrange 4 objects= 24

Required probability = 9/24 =3/8

I am not familiar with derangement, but it does look like a useful thing to know. However, there is a way to argue it based on Inclusion - Exclusion principle.

The total number of ways to arrange the objects is 4! Now of these let us exclude the cases when 1 is in the 1st place and then the number of cases when 2 is in the second place and so on. That gives 4X3! = 4!

But in doing so, we have counted twice the number of cases when the pairs 6 pairs (1,2), (1,3) ) (2,3) etc. are in their respective positions and so we must include them back and that is 6X2 = 12 cases

Now, again we have to exclude the cases when the 4 triplets (1,2,3), (2,3,4) etc. are in their respective places and that is 4.

And finally include the one case when all four are in their respective positions.

So that gives number of favourable cases = 4!-4!+12-4+1 = 9

Hence probability = 9/24 = 3/8

Suppose we have  a set of n objects A = {a₁, a₂, . . ., a_n}. Then, we have a one-to-one correspondence between the set of these objects and the set of integers S = {1, 2, . . ., n}. So if we talk of the permutations of the set A, we can simply talk of the permutations of set S. Any permutation of the integers can be thought of as a function from the set S to itself. In fact, it IS a bijetive function. For example, if we consider the integers 1, 2, 3, then its permutations are 123, 132, 231, 213, 321, 312. Take any one permutation, say 213. Then, this actually is the image of 123 under the function f so that f(1)=2, f(2)=1, f(3)=3.

Now, if there is an element k in the set S, so that f(k)=k, then k is called the fixed point of the permutation f. And, if a permutation has no fixed points, it is called a derangement. It can be easily shown (using PIE, for example) than the number of derangements D_n of the set S is