[Guest]

All the 7! = 5040 distinct permutations of the positive integer 1234567, including itself, are arranged in accordance with ascending order of magnitude.

Determine the 2010^th number in the above arrangement.

[Guest]

1234567 = n(1)

1324567 = n(2)

...

xxxxxxx = n(2010)

...

7564321 = n(5039)

7654321 = n(5040)

If this is right I'm sure I'll eventually figure it out but I have to clean up and go to work(stupid work).

Straight guessing from the obvious eventual pattern. It should be between 3576421 and 4123567.

All the 7! = 5040 distinct permutations of the positive integer 1234567, including itself, are arranged in accordance with ascending order of magnitude.

Determine the 2010^th number in the above arrangement.

Edited June 2, 2010 by PVRoot

[Guest]

1234567 = n(1)

1324567 = n(2)

...

xxxxxxx = n(2010)

...

7564321 = n(5039)

7654321 = n(5040)

If this is right I'm sure I'll eventually figure it out but I have to clean up and go to work(stupid work).

Straight guessing from the obvious eventual pattern. It should be between 34xxxx and 43xxxx.

Actually:

n(1)=1234567

n(2)=1234576

n(3)=1234657

n(4)=1234675

I gotta get to sleep so I might come back and take a jab at this later as well, wanna race?

Edited June 2, 2010 by Anza Power

[Guest]

Wait a second actually this doesn't take much thinking at all...

The answer is 1...

The reason for this is the 2010th number is actually the first number of the 288th permutation, according to how the numbers are arranged the first 720 (6!) permutation are 1###### (with ###### being all permutation for 234567)

[superprismatic]

We count the permutations using the following

table:

perms starting with Number Cumulative
1 720 720
2 720 1440
31 120 1560
32 120 1680
34 120 1800
35 120 1920
361 24 1944
362 24 1968
364 24 1992
3651 6 1998
3652 6 2004
3654 6 2010

So, it's the last permutation starting with 3654.

That would be 3654721.

[Guest]

We count the permutations using the following

table:

perms starting with     Number      Cumulative

        1                720           720

        2                720          1440

       31                120          1560

       32                120          1680

       34                120          1800

       35                120          1920

      361                 24          1944

      362                 24          1968

      364                 24          1992

     3651                  6          1998

     3652                  6          2004

     3654                  6          2010

So, it's the last permutation starting with 3654.

That would be 3654721.

Well my first guess was close I'd have gotten there.

[Guest]

2010 = (2*6!) + (4*5!) + (3*4!) + (3*3!)

Hence no. is 3654127

[Guest]

Wait is it the 2010th permutation or the 2010th digit if you put all the permutations one alongside the other in a string?

[Guest]

Wait is it the 2010th permutation or the 2010th digit if you put all the permutations one alongside the other in a string?

It is the 2010th permutation and, not the 2010th digit, if all the permutations are placed one alongside the other in a string.