derrangements

number of ways in whici n letters can be placed in n directed envolopes so that no letter goes into its own envelope is

n! [ 1/2!-1/3! +1/4!-1/5!..........+(-1)ⁿ1/n!]

here n=4 is tha case, so substitute n=4.

now dont say i am a formula mugger, this is quite a general one and have learnt after solving many such problems. and ya its not always given as letters, so relate with the question and u can find use of this formula in other cases also.

ok, so i will try to put up something........

ABOVE ALL A1,A2,A3......Am are finite sets, forgot to mention that.

MY APPROACH TO THIS PROBLEM WAS AS BELOW PLS POINT OUT ERROR: TOTAL NO. OF WAYS . = 4! ........

NOW LET SETS A BE NO. OF WAYS IN WHICH LETTER 1 GOES TO RITE PLACCE ..B FOR LETTER 2 .....

AND SO ON TILL D ...........

NOW I FOUND n(A U B U C U D) = n(A) + n(B)...+n(D) - n(A∩B).......+n(A∩B∩C)......-n(A∩B∩C∩D) ...................

THEN I DID 4! - n(A U B U C U D) as we dont want anything goin to rite place........

but my answer is not correct explain the flaw in mah method [2][2]
{edited}

now, we can see

if A1,A2,A3............Am are subsets of AA containing N elements, then

n(A₁`∩A<sub>2</sub>`∩...........∩A_m`)

= N - Σ n(A_i) +Σ n(A_i∩A_j) - Σ n(A_i∩A_j∩A_k)+......
i 1≤i<j≤m 1≤i<j<k≤m

there kindly note that in the formula i bolded, i have used compliments of A1,A2,A3..........

now this above post is what leads us to DERRANGEMENTS.