                    ABCD
                    CABD
                    ACDB
                    DACB
                    BCDA
                    ACBD
                    ADCB
                    CDAB
                    DABC
                    BCAD
                    CADB
                    CDBA
                    CBAD
                    ABDC
                    ADBC
                    BDCA
                    DCBA
                    BACD
                    BADC
                    BDAC
                    CBDA
                    DBCA
                    DCAB

Listed above are   all-but-one   of the permutations of the symbols   A,   B,   C,   and   D,   except   for one permutation that's   not   listed.



Task

Find that missing permutation.



Methods

 Obvious method: 
        enumerate all permutations of   A,  B,  C,  and  D,  
        and then look for the missing permutation. 

 alternate method:
        Hint:  if all permutations were shown above,  how many 
        times would  A  appear in each position?     
        What is the  parity  of this number?

 another alternate method:
        Hint:  if you add up the letter values of each column, 
        does a missing letter   A,  B,  C,  and  D   from each
        column cause the total value for each column to be unique?



Related task

   Permutations)


#include <stdio.h>
 
#define N 4
const char *perms[] = {
	"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",
	"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",
	"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB",
};
 
int main()
{
	int i, j, n, cnt[N];
	char miss[N];
 
	for (n = i = 1; i < N; i++) n *= i; /* n = (N-1)!, # of occurrence */
 
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) cnt[j] = 0;
 
		/* count how many times each letter occur at position i */
		for (j = 0; j < sizeof(perms)/sizeof(const char*); j++)
			cnt[perms[j][i] - 'A']++;
 
		/* letter not occurring (N-1)! times is the missing one */
		for (j = 0; j < N && cnt[j] == n; j++);
 
		miss[i] = j + 'A';
	}
	printf("Missing: %.*s\n", N, miss);
 
	return 0;
 
}