By the theory of permutation, arranging the letter in alphabetical order, total no. of letters made from D =4!/2!=12,same for E &I,total no. of words made from D,E,I=12*3=36.
Now fixing R in the first place,no. of letters having RD and RE=3!*2=12.
Total no. of words before RIDER=36+12=48,
.'.RIDER will be at 49th place.
Hence, option 6) is the right answer.

Word starting with D has 4!/2!=12 combinations
Word starting with E has 4!/2!=12 combinations
Word starting with I has 4!/2!=12 combinations
Word starting with RD has 3!=6 combinations
Word starting with RE has 3!=6 combinations
Then the word comes RIDER
So, it's rank will be=12+12+12+6+6+1=49
So, option 6)49 is the right answer

D_ _ _ _ it will be starting from D.Now we have to fill up 4 positions after D.so it would be (!4/!2).I'm dividing factorial 2 because R is given two times.
So D_ _ _ _=!4/!2=>12.
Same thing goes for E_ _ _ _=>12.
Same thing goes for I _ _ _ _=>12.
Now when it comes on R _ _ _ _. for the 2nd position it would be like this.R D E I R(for the first word starting from R).R D _ _ _.now for remaining three condition !3=6
When R E _ _ _. then remaining condition will be !3=6.
after this next word will be R I D E R according to alphabet.
Total=12+12+12+6+6+1=>49
So the rank is 49.

3/2!*4! + 2*3! + 0*2! + 0*1! + 0*1 + 1 = 49

R I D E R
4 3 1 2 4

3/2! 2 0 0 0
4! 3! 2! 1! 0!

rank of the word rider is 3*2!*4!+2*3!+1=36+12+1=49

option 6 is correct

d---- 12
e---- 12
i ---- 12
rd--- 6
re--- 6
rider +1
49

FOR D=4!/2!=12
FOR E=4!/2!=12
FOR I=4!/2!=12
NOW R D _ _ _ =3!=6
THEN R E _ _ _ =3!=6
AND R I D E R=1
TOTAL=12+12+12+6+6+1=49

STARTING WITH DICTIONARY ALPHABETS
DE*** ----1 TO 3 WORD (SINCE R IS IDENTICAL HERE THTS WHY ONLY 3 PERMUTATIONS POSSIBLE)
DI***--- 4 TO 6TH END WORD
DR***--- 7 TO 12TH WORD
SIMILARLY ER***---- 13 TO 24TH WORD
IR***---25TH TO 36TH WORD
RD***---37 TO 42ND WORD
RE***---43 TO 48TH WORD
RIDER----49TH WORD
HENCE OPTION 6.)

3*(4!/2)+(2*3!)+1

Words will be in alphabetical order....
Words starting with d = 4!/2=12
Words starting with e=12
Words starting with I=12
Words starting with rd=3!
Words with re=3!=6
Then acc to alphabetical order rider comes...
So rank of word rider =12+12+12+6+6+1=49

Rank = 3* 4!/2! + 2*3! + 1 = 36+12+1 = 49. Ans.

Words starting from d are 12
Words from e 12
Words from i 12
Words from RD are 6
Words from RE are 6
And the last 49th word is RIDER

the letter starting with d is 4!/2!---12
similarly starting with e is 12
for i also 12
and for starting with rd....6
and for re -----------6
then rider comes...
so total =36+12+1=49

D E I R R
before R we have 3 letters so 3*4!/2!=36
befor I we have 2 letters so 2*3!=12
then before D and E we dnt have any letters so
36+12+1=49

Acc to alphabetical order......
D,E,I,R,R
(4*3*2*3/2) + (3*2) + 1 = 49

arrage in alphabatic order
D E I R R
now
D=>4!/2!=12
E=>4!/2!=12
I=>4!/2!=12
RD=>3!=6
RE=>3!=6
RIDER=1
total=12+12+12+6+6+1=49 ans .

option 6

it is ans
(3/2!)*4! =2*3! +1=49

Number of words which starts from D = 12
Number of words which starts from E = 12
Number of words which starts from I = 12
Number of words which starts from RD = 12
After RD, words starts from RI and first word is RIDER
so position= 12+12+12+12+1 = 49

The answer is 49.
No. of words starting from D=4!/2!
No. of words starting from E=4!/2!
No. of words starting from I=4!/2!
No. of words starting from RD=3!
No. of words starting from RE=3!
The next word will be RIDER.
Thus Rank of RIDERh in the dictionary will be=3*(4!/2!)+2*3!+1=49. ANS.

36+6+6+1=49

ANSWER SHOULD BE : 49


// CODE TO PROVE IT

#include
#include

long fact(int n)
{
long f = 1;
while (n > 0)
{
f *= n;
n--;
}
return f;
}

int find_rank(char s[], int len)
{
long rank = 1;
int i, j, k, l, fr[26], f; //fr is frequency
char sorted[100], cur, t, pre;

for (i = 0; i < len -1 ; ++i)
{
cur = s[i];
strcpy(sorted, s + i);
l = len - i;
memset(fr, 0, sizeof(fr));
fr[sorted[0] - 'a']++;
for (j = 1; j < l; ++j)
{
t = sorted[j];
fr[t-'a']++;
for (k = j-1; k >=0 && sorted[k] > t; --k)
sorted[k+1] = sorted[k];
sorted[k+1] = t;
}
pre = 1;
for (k = 0; sorted[k] != cur ;++k)
{
if (sorted[k] == pre)
continue;
pre = sorted[k];

f = fact(l - 1);
fr[pre - 'a']--;
for (j = 0; j < 26; ++j)
if (fr[j])
f /= fact(fr[j]);
rank += f;
fr[pre - 'a']++;
}
}
return rank;
}

int main()
{
char s[100];

scanf("%s", s);
printf("%d", find_rank(s, strlen(s)));
return 0;
}