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A permutation puzzle (Posted on 2006-09-20) Difficulty: 3 of 5
Determine the number of permutations (p1, p2,...p7) of 1,2, ...7; such that for all k 1≤k≤6, (p1, p2,... pk) is not a permutation of (1,2, ...k); i.e., p1≠1; (p1, p2) is not a permutation of (1,2), etc.

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

No Solution Yet Submitted by K Sengupta    
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solution? | Comment 2 of 4 |

For the first 6 numbers not to be a permutation of 1..6, then one of the numbers must be 7. So, the question is how many permutations are there where 7 isn't in the last position?

We could determine this by

total permutations - permutations ending with 7
(The permutations ending with 7 is the number of permutations of 1..6.)

7! - 6!

= 5040 - 720

= 4320

I think this generalizes to
N! - (p!.N-p!)

Where N is the maximum length of the sequence, and p is the length of the prefix that should not contain numbers  1..p.

For 1<=k<6, this gives N=7, p=5,
so
7! - (5!.2!)
= 5040 - (120 x 2)
= 4800

I'm not often on the ball with this kind of stuff so half expecting this to be wrong!


  Posted by bumble on 2006-09-20 19:16:35
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