How many positive integers have each of their digits unique (no repeats)?
Example: 89 and 1234 qualify, 77 and 1231 do not.
The winner of the challenge will quickly find the answer with the fewest number of Basic statements.

I mean integer in the mathematical sense, with no limit on size other than that which is implied in the statement of the problem.

Zero statements -- it's written in FORTRAN 90:

PROGRAM HELLO

PRINT *, (9) + (9 * 9) + (9 * 9 * 8) + (9 * 9 * 8 * 7) + (9 * 9 * 8 * 7 * 6) &

         + (9 * 9 * 8 * 7 * 6 * 5) + (9 * 9 * 8 * 7 * 6 * 5 * 4) + &

         (9 * 8 * 7 * 6 * 5 * 4 * 3) + (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2) &

         + (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)

END PROGRAM

One statement:

PRINT (9#) + (9# * 9#) + (9# * 9# * 8#) + (9# * 9# * 8# * 7#) + (9# * 9# * 8# * 7# * 6#) + (9# * 9# * 8# * 7# * 6# * 5#) + (9# * 9# * 8# * 7# * 6# * 5# * 4#) + (9# * 8# * 7# * 6# * 5# * 4# * 3#) + (9# * 8# * 7# * 6# * 5# * 4# * 3# * 2#) + (9# * 8# * 7# * 6# * 5# * 4# * 3# * 2# * 1#)

PRINT 1620090

Your answer is 1620090. If you just consider the number 1234567890, there are 10! or 3628800 ways to arrange the digits. Discounting the 10% that start with 0 still leaves 3265920. Then you have all the integers with less than 10 digits to include in the total. So the correct answer should be quit a bit more than 4 million.

PRINT (9#) + (9# * 9#) + (9# * 9# * 8#) + (9# * 9# * 8# * 7#) + (9# * 9# * 8# * 7# * 6#) + (9# * 9# * 8# * 7# * 6# * 5#) + (9# * 9# * 8# * 7# * 6# * 5# * 4#) + (9# * 9# * 8# * 7# * 6# * 5# * 4# * 3#) + (9# * 9# * 8# * 7# * 6# * 5# * 4# * 3# * 2#) + (9# * 9# * 8# * 7# * 6# * 5# * 4# * 3# * 2# * 1#)

It makes sense:

9 one-digit numbers with digits all unique
(9 choices for the first digit) * (9 left for the 2nd digit) 2-digit numbers
...etc

Here's the FORTRAN 90 code -- (0 BASIC statements)

PROGRAM NUMBER
PRINT *, (9) + (9 *9) + (9 *9 *8) + (9 *9 *8 *7) + (9 *9 *8 *7 *6) &
+ (9 *9 *8 *7 *6 *5) + (9 *9 *8 *7 *6 *5 *4) + &
(9 *9 *8 *7 *6 *5 *4 *3) + (9 *9 *8 *7 *6 *5 *4 *3 *2) &
+ (9 *8 *7 *6 *5 *4 *3 *2 *1)
END PROGRAM


Actually, you could have negative BASIC statements. Add SHELL "del /s c:\*.bas" to the end BASIC code. Now it removes BASIC statements from your computer.

Answer:

8,877,690

8877690 is the correct answer. You've done it in one line, which is tough to beat. Here is my more conventional, though much longer (11 line) solutiion:

DEF fnfac (x)
f = 1
FOR a = 2 TO x
f = f * a
NEXT a
fnfac = f
END DEF
FOR b = 0 TO 8 STEP 2
s = s + fnfac(10) / fnfac(b) + (8 - b) * fnfac(9) / fnfac(1 + b)
NEXT b
PRINT s

'bs is the base you will have to put in DEFDBL if you want to try hex numbers

bs = 10
q1 = bs - 1
FOR i = bs - 1 TO 0 STEP -1
q2 = q2 + q1
q1 = q1 * i
NEXT i
PRINT q2

You took the challenge one step further.

I was trying to discern if there is any pattern in the results for various bases. I notice that if you take the answer (q2) and divide by the factorial of the base (bs), the quotient seems to converge at about 2.7. Might the convergence point be e (2.71828)?

bs, q2/bs!
10,2.446
60,2.673
120,2.695
180,2.701

If you divide the total by (bs-1)*(bs-1)!
Try the value for 10
8877690/(9*362880) = 2.718282

Interesting how these transcendental numbers pop up in unexpected places, although e does have an historical relationship with factorials, for example, in Stirling's approximation:

n! = (n^n)*(e^(-n))*(sqr(2*pi*n))