PREMIUM PERMUTATION.......!!!!

For each positive integer k, 1<=k<=5,let Nk denote the number of permutations (p1 , p2 , ......, p6) such that p1 not equal to 1,(p1 , p2) is not a permutation of (1,2),.....,(p1 , p2 , ......, pk) is not a permutation of (1,2,3,....k).we require to find N5 . we shall start with N1 . since out of 6! permutations of (1,2,3,....,6), there are 5! permutations having 1 in the first place, so so we have to find the number of permutations left after removing from set of all permutations of (1,2,3,...,6), the ones that begins with 1. consequently N1=6!-5!=600. to get N2 we remove those permutations in which (p1 , p2) is a permutation of (1,2).since all permutations of the type (1,2,p3,....p6) have already been removed,we have to again remove the permutation of the type (2,1,p3, . . . .,p6) .the number of such permutation being 4!,we get N2=600-4!=576.having removed the permutations beginning with (1,2),we should now remove those beginning with (1,2,3).but corresponding to first two places (1,2) & (2,1), we have removed all the permutations .so we now, remove the permutations with 1st three places (3,2,1),(3,1,2),(2,3,1).note that the 1st three positions being 1,2,3 is included in the permutations beginning with 1. for each of these three arrangements ,there are 3! ways arranging 4th , 5th & 6th places and so N3=N2 - 3 3!=576-18=558.similarly for N4 first four elements are a permutation of (1,2,3,4).there are (3! +2! +2! +1 +1) 2=24 permutation in all to be removed .so N4= 558-24=534.to get N5 ,we have to remove permutations (p1 , p2 , p3,...,p6) in which (p1,p2,.....,p5)is a permutation of (1,2,3,4) and which have not been removed .there are 4! such permutations for which p1=5,18 with p2=5,16 with p3=5 and 13 with p4 =5,so in all there are 71 such permutations.thus N5=534-71=(24+18+16+13)=463 is the required number of permutations.