Six cards and six envelopes are numbered 1,2,3,4,5,6 and cards are to be place 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 is?
My approach: Suppose '1' is fixed at 2nd position then there will be 120 arrangements all total including all the bad permutations and dearrangments.
Now, i need to count only bad permutations which are not dearrangements.
Suppose set A3 contains arrangements which have third letter at 3rd envelope , A4= 4th letter at 4th envelope ...
then,
Cardinality of |A3 Union A4 union A5 union A6|= the formula is summation of ((-1)^(k+1) )* (n!/k!) where k starts from 1 and ends at n where n is the no of sets right?. or n represents the last set i.e 6 ?
which one
If former, 4! [ 1/1! -1/2! + 1/3! - 1/4!] = 15 coming as n=4 there are 4 sets considered .
if Latter, then 6! * [ 1/3! -1/4! + 1/5! - 1/6!] = 95. ( no of bad permutations )
My idea is to find the no of bad permutations and subtract it from 120 to get answer. Where am i wrong?
My approach: Suppose '1' is fixed at 2nd position then there will be 120 arrangements all total including all the bad permutations and dearrangments.
Now, i need to count only bad permutations which are not dearrangements.
Suppose set A3 contains arrangements which have third letter at 3rd envelope , A4= 4th letter at 4th envelope ...
then,
Cardinality of |A3 Union A4 union A5 union A6|= the formula is summation of ((-1)^(k+1) )* (n!/k!) where k starts from 1 and ends at n where n is the no of sets right?. or n represents the last set i.e 6 ?
which one
If former, 4! [ 1/1! -1/2! + 1/3! - 1/4!] = 15 coming as n=4 there are 4 sets considered .
if Latter, then 6! * [ 1/3! -1/4! + 1/5! - 1/6!] = 95. ( no of bad permutations )
My idea is to find the no of bad permutations and subtract it from 120 to get answer. Where am i wrong?