| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Gee, there sure have been alot of problems submitted on the subject of permutations of the digits 0123456789. I just thought I'd rise to the occasion. Here's one I just made up. Consider a long plank to be used as a see-saw. The plank is labeled evenly from one end to the other with all the integers from 0 to 9999999999. Now, put many "seats" on the see-saw, with seats located at all the integers formed of the permutations of the ten decimal digits. So, seats are located at: 0123456789, 0123456798, . . . 9876543210, 9876543201 There are a total of 10-factorial (10!) seats. The question: At exactly what point should the fulcrum be placed so the see-saw balances with seats and no riders ?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 247.1 | KOBAL::GILBERT | Mon Mar 25 1985 12:26 | 12 | ||
I'll assume the plank is weightless, but the seats are not. The problem is then equivalent to finding the average of all permutations of the digits 0123456789. To find the average, we sum the 10! permutations, and divide by 10!. Note that in each column of the summation, each of the ten digits occurs equally often. So, Sum of the permutations = 10! * (1111111111) * (0+1+2+3+4+5+6+7+8+9)/10 = 10! * (10**10-1)/9 * 9/2 = 10! * (10**10-1) / 2 Average of the permutations = (10**10 - 1)/2 Thus, the fulcrum should be placed in the middle (at 9999999999/2), and it turns out that the assumption of a weightless plank is unnecessary. | |||||