Determine the number of permutations (p

_{1}, p

_{2},...p

_{7}) of 1,2, ...7; such that for all k 1≤k≤6, (p

_{1}, p

_{2},... p

_{k})
is not a permutation of (1,2, ...k); i.e., p

_{1}≠1; (p

_{1}, p

_{2}) is not a permutation of (1,2), etc.

What would be the answer if we specify 1≤k<6 instead?

Total permutations = 7! = 5040

1 <= k <= 1

The number of permutations which start with 1 are 6! = 720.

So p1 not a permutation of (1) = 5040 - 720 = 4320

1 <= k <= 2

The number of addtional permutations where (p1,p2) is a permutation of (1,2) but p1 is not a permutation of 1:

Necessarly p2 must be 1 and p1 must be 2. So additional permutations = 5! = 120.

The number of permutations where p

_{1}≠1 and (p

_{1}, p

_{2}) is not a permutation of (1,2) = 4320 - 120 = 4200

1 <= k <= 3

The number of addtional permutations where (p1,p2,p3) is a permutation of (1,2,3) but p1 is not a permutation of 1 and (p

_{1}, p

_{2}) is not a permutation of (1,2):

Necessarly p3 must be 1 or 2 and (p1,p2) must be a permutation of the
the other 2. And (p4,p5,p6,p7) is a permutation of (4,5,6,7) So
additional permutations = 2*2!*4! =96

.

The number of permutations where p

_{1}≠1 and (p

_{1}, p

_{2}) is not a permutation of (1,2) and (p1,p2,p3) is not a permutation of (1,2,3) = 4320 - 120 = 4200 - 96 = 4104

1 <= k <= 4

Similarly, 4104 - 3*3!*3! = 4014 - 108 = 3996

1 <= k <= 5

Similarly, 3996 - 4*4!*2! = 3996 - 96 = 3900

This is my answer to the 2nd question.

Hope I didn't make a mistake

1 <= k <= 6

Similarly, 3900 - 5*5!*1! = 3900 - 600 = 3300

This is my answer to the 1st question.

Hope I didn't make a mistake