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Q: Numeric Permutations ( Answered 5 out of 5 stars,   3 Comments )
Subject: Numeric Permutations
Category: Science > Math
Asked by: da_maury-ga
List Price: $5.00
Posted: 01 Nov 2002 13:51 PST
Expires: 01 Dec 2002 13:51 PST
Question ID: 95825
Given a sequence from 1 to 8, what formula will calculate each unique
combination using all 8 values.  I believe there are 256 unique

For example: Seq 1)  1,2,3,4,5,6,7,8
Seq 2) 2,1,3,4,5,6,7,8

Although I can do this by hand, it is not efficient.

The correct answer should be adaptable to a computer routine,
preferably in C++ or VB.

Clarification of Question by da_maury-ga on 01 Nov 2002 14:12 PST
I do not want the individual values re-used in a given sequence
(therefore, the sequence 1,1,2,3,4,5,6,7 is invalid.)  Each sequence
must use all values once and only once.

Subject: Re: Numeric Permutations
Answered By: till-ga on 01 Nov 2002 14:45 PST
Rated:5 out of 5 stars
First of all the definition for your problem:
A permutation is a way to order a set of things. For example, if your
set is the letters in the word WHO, then one other ordering would be
WOH. Here are all the possible orderings of the letters in the word
There are 6 different ways to order the letters in the word WHO.
How about the word "STOP"? Well, here they are:

   STOP     STPO     SOTP     SOPT     SPTO     SPOT <- starts with
   TSOP     TSPO     TOSP     TOPS     TPSO     TPOS <- starts with
   OSTP     OSPT     OTSP     OTPS     OPST     OPTS <- starts with
   PSTO     PSOT     PTSO     PTOS     POST     POTS <- starts with

There are 24 ways to order the letters in "STOP". Is there a general
rule here? Fortunately, yes. Here's the rule for "STOP":
There are 4 ways to pick the first letter. 
After you pick the first letter there are 3 ways to pick the second
After you pick the first 2 letters, there are 2 ways to pick the third
After picking the first 3 letters, there is only 1 letter left to
So the number of ways to order the letters in "STOP" is 4 x 3 x 2 x 1
= 24 ways!"
( )

So your first guess of 256 is wrong. The rule applied to your problem
The number of permutations of your sequence of eight numbers is:

8! = 40.230
The following c routine will output all combinations as desired:

/* C program for distribution from the Combinatorial Object Server. 
/* Generate permutations in lexicographic order. This is            
/* the same version used in the book "Combinatorial Generation" by  
/* Frank Ruskey.                                                    
/* The program can be modified, translated to other languages, etc.,
/* so long as proper acknowledgement is given (author and source).  
/* Programmer: Frank Ruskey 19??, Joe Sawada, 1997.                 
/* The latest version of this program may be found at the site      

 * Modified to print permutation to file.
 * may 1st, 2002 by edmond

#include <stdio.h>
#include <stdlib.h>

int  n;
int  a[100];   /* The permutation */
char filename[30];
FILE *infile;

void PrintPerm() {

	int i;
	for (i=1; i <= n; i++){
		fprintf( infile, "%d ", a[i] );
		printf ( "%d ", a[i]);
	fprintf(infile, "\n");
	printf ("\n");

void swap(int i, int j) {

	int temp;
	temp = a[i];
	a[i] = a[j];
	a[j] = temp;

int Next() {

	int k,j,r,s;

	k = n-1; 
	while (a[k] > a[k+1]) k--;
	if (k == 0) return(0);
	else {
		j = n;
		while (a[k] > a[j]) j--;
		r = n; s = k+1;
		while (r>s) {
			r--; s++;

int main () {

	int i;

	printf( "Enter n: " );  scanf( "%d", &n );

	if (n<=0) exit(1); 

	printf ( "Enter filename: " ); scanf( " %s", filename);

	infile = fopen( filename, "w");

	if( !infile ){
		printf("Error opening file to write\n");
		return 0;

	printf( "\n" );
	for (i=0; i<=n; ++i) {
		a[i] = i;
	while (Next());
	printf( "\n" );

	printf("Press Enter to Continue...");
	fflush( stdin );
	return 0;
( )

Search Strategy:
( ://
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da_maury-ga rated this answer:5 out of 5 stars
Excellent and thorough response! By the way, thank you to both
researchers for the correction to my underestimate.  I'll be back with

Subject: Re: Numeric Permutations
From: mcfly-ga on 01 Nov 2002 14:09 PST
Just as a note to any researcher who attempts this question, after a
couple of minutes thinking about a possible solution, I think there's
actually 8!=40320 possible permutations.
Subject: Re: Numeric Permutations
From: kdb003-ga on 01 Nov 2002 14:56 PST
I just want to point out that the calculation of 8!=40230 is not
correct in the researchers response, and is correct(8!=40320) in the
previous comment.  Probably just a typo.
Subject: Re: Numeric Permutations
From: till-ga on 02 Nov 2002 06:20 PST
Thank you kdb003-ga for your hint at my typo. Of course you are right:
Sorry for that da_maury-ga  and thanks for still giving the high rating.


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