### Dictionary Rank of a Word | Permutations & Combinations

PERMUTATIONS & COMBINATIONS

Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc.

# Dictionary Rank of a Word

Method of finding the Rank (Dictionary Order) of the word “R A I N”
Given word: R A I N
Total letters = 4
Letters in alphabetical order: A, I, N, R
No. of words formed starting with A = 3! = 6
No. of words formed starting with I = 3! = 6
No. of words formed starting with N = 3! = 6
After N there is R which is required
R ----- Required
A ---- Required
I ---- Required
N ---- Required
RAIN ----- 1 word

 RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19

## Method of finding the Rank (Dictionary Order) of the word  “F A T H E R”

Given word is :  "F A T H E R"

In alphabetical order: A, E, F, H, R, T

Words beginning with A = 5! = 120

Words beginning with E = 5! = 120

Words beginning with F… Required

Alphabet left A, E, H, R, T

Words beginning with FA… Required

Alphabet left  E, H, R, T

Words beginning with FAE = 3! = 6

Words beginning with FAH = 3! = 6

Words beginning with FAR = 3! = 6

Words beginning with FAT…..Required

Alphabet left  E, H, R

Words beginning with FATE = 2! = 2

Words beginning with FATH ….. Required

Alphabet left  E, R

Words beginning with FATHE ….. Required

Alphabet left  R

Words beginning with FATHER = 1 word

Rank of FATHER = 120 + 120 + 6 + 6 + 6 + 2 + 1 = 261

 RANK OF THE WORD “F A T H E R” A ….. = 5!    = 120 A ….. = 5!    = 120 FAE …. = 3! = 6 FAE …. = 3! = 6 FAE …. = 3! = 6 FATE = 2!    = 2 F A T H E R  = 1 120 120 6 6 6 2 1 TOTAL 261 Rank of “F A T H E R” is 261

## Method of finding the Rank (Dictionary Order) of the word  “COLLEGE”

Letters in the ascending order: C, E, E, G, L, L, O

Letter ‘C’ is required so we left with E, E, G, L, L, O

Now number of words formed starting with C
C E _ _ _ _ _ = 5!/2! = 60 words
Note: Here letter ‘L’ repeat twice so we divide by 2!
C G _ _ _ _ _ = 5!/(2!*2!) = 30 words
Note: Here letters ‘L’ and ‘E’ repeat twice so we divide by 2!*2!
C L _ _ _ _ _ = 5!/2! = 60 words
C O -------- = Required

C O is required so we left with E, E, G, L, L

Now number of words formed starting with C O
C O E _ _ _ _ = 4!/2! = 12 words
C O G _ _ _ _ = 4!/(2!*2!) = 6 words
C O L ----- = Required

C O L is required so we left with E, E, G, L
Now number of words formed starting with C O L
C O L E _ _ _ = 3! = 6words
C O L G _ _ _ = 3!/2! = 3 words
C O L L ----- = Required
C O L L E ----- = Required
C O L L E is required so we left with E, G

Now number of words formed starting with C O L L E
C O L L E E _ = 1! = 1 word
C O L L E G E = 1 word
TOTAL = 60 + 30 + 60 + 12 + 6 + 6 + 3 + 1 + 1 = 179
Summery

 C O L L E G E C E _ _ _ _ _  = 5!/2! C G _ _ _ _ _ = 5!/(2!*2!) C L _ _ _ _ _ = 5!/2! C O E _ _ _ _ = 4!/2! C O G _ _ _ _ = 4!/(2!*2!) C O L E _ _ _ = 3! C O L G _ _ = 3!/2! C O L L E E _ = 1! C O L L E G E 60 30 60 12 6 6 3 1 1 Total 179 Rank of “C O L L E G E” is 179

Method of finding the Rank (Dictionary Order) of the word “SUCCESS” Letters in the alphabetical order: C, C, E, S, S, S, U
First letter which is required is ‘S’
Number of words starting with C = 6!/3! = 120 ( permutations of SUCESS )
Number of words starting with E = 6!/(3!*2!) = 60 ( permutations of SUCCSS )
After E the letter is S
Since S is the first letter of the given word so it is Required
S ------ = Required
Now we have letters: C, C, E, S, S, U
No. of words formed by fixing ‘S’ at first position

Number of words formed starting with SC = 5!/2! = 60 ( permutations of UCESS )
Number of words formed starting with SE = 5!/(2!*2!)= 30 ( permutations of UCCSS )
Number of words formed starting with SS = 5!/2! = 60 = ( permutations of UCCES )
After S the letter is U
SU ----- = Required
SUC ----- = Required
SUCC------ = Required
SUCCE------ = Required
SUCCS------ = Required
SUCCESS------ = Required = 1 word
Rank of “SUCCESS” = 120 + 60 + 60 + 30 + 60 + 1 = 331

 RANK OF THE WORD “S U C C E S S” E = 6!/(3!*2!)  C = 6!/3!  SC = 5!/2!  SE = 5!/(2!*2!) SS = 5!/2! SU….SUC….SUCC….SUCCE…. SUCCES…SUCCESS 60 120 60 30 60   1 TOTAL 331 Rank of “S U C C E S S” is 331

## Method of finding the Rank (Dictionary Order) of the word  “S O C C E R”

Given word: S O C C E R
Total letters = 6
Letters in alphabetical order: C, C, E, O, R, S
No of words starting with C = 5! = 120 (Permutations of C, E, O, R, S)
No of words starting with E = 5!/2! = 60 (Permutations of C, C, O, R, S)
No of words starting with O = 5!/2! = 60 (Permutations of C, C, E, R, S)
No of words starting with R = 5!/2! = 60 (Permutations of C, C, E, O, S)
After R Next letter is S
Since S is the first letter of the given number so it is Required
S ------ = Required
Now we have letters: CCEOR
No. of words starting with "S"
o of words starting with SC = 4! = 24 (Permutations of CEOR)
No of words starting with SE = 4!/2! =12 (Permutations of CCOR)
After E Next letter is O
O---- = Required
Now we have letters: CCER
No of words starting with SO….
SOC----=Required
SOCC-----=Required
SOCCE ---- = Required
SOCCER ---- = Required = I word

 RANK OF THE WORD “S O C C E R” C = 5! = 120  E = 5!/2! = 60 O = 5!/2! = 60  R = 5!/2! = 60  SC = 4! = 24 SE = 4!/2! =12 SO....SOC.....SOCC.....SOCCE.....SOCCER 120 60 60 60 24 12 1 TOTAL 337 Rank of “S O C C E R” is 337

Method of finding the Rank(Dictionary Order) of the word  “C O M P U T E R”
Given word: C O M P U T E R
Total letters = 8

Letters in alphabetical order: C, E, M, O, P, R, T, U
Since C is the first letter of the given number so it is Required
C ------ = Required
Now we have letters: E, M, O, P, R, T, U
After C there is E
No. of words formed starting with C
CE -------- = 6! = 720
CM -------- = 6! = 720
CO -------- = Required
Now we left with: E, M, P, R, T, U
Now number of words formed starting with CO
C O E ------- = 5!
C O M ------ = Required
We left with: E, P, R, T, U
Now number of words formed starting with C O M
C O M E ---- = 4!
C O M P ---- = Required

Now we left with: E, R, T, U
Now number of words formed starting with C O M P
C O M P E ----- =3!
C O M P R ---- = 3!
C O M P T ---- = 3!
C O M P U --- = Required
Now we left with: E, R, T
Now number of words formed starting with C O M P U
C O M P U E --- = 2!
C O M P U R --- = 2!
C O M P U T ---- = Required
Now we left with: E, R
Now number of words formed starting with C O M P U T
COMPUTER = Required = 1 word
Hence Rank of the word “C O M P U T E R” is
= 720+720+120+24+6+6+6+2+2+1= 1607

## COMPUTER

CE ---------------  6!

CM -------------- 6!

COE ------------- 5!

COME ---------- 4!

COMPE----------3!

COMPR -------- 3!

COMPT -------- 3!

COMPUE ------ 2!

COMPUR -----  2!

COMPUTER ----1

720

720

120

24

6

6

6

2

2

1

Total

1607

Rank of “COMPUTER” is 1607

# SECOND METHOD OF FINDING THE RANK OR DICTIONARY ORDER OF A WORD

Let us explain this method by taking an example of the word "SCHOOL"
1) Write the given word in capital letters.

2) Write the numerical values (Natural numbers) on the letters in alphabetically
3) Now count the total letters in the word. For example no. of letters in the word SCHOOL is 6.
4) Subtract 1 from the total letters of the word (6-1=5)
5) Start from the left hand side: start writing 5!, 4!, 3!, 2!, 1!, 0! below the letters of the given word.
6) Start from the extreme  LHS and write the number of letters on the Right Hand Side of the selected alphabet which have less numerical value  than the value than the selected alphabet and divide with the number of repeating alphabets to the RHS of the selected alphabet.

Eg; Numerical value on "S" is 5
LHS of letter "S" there are 5 letter with  numerical values (1, 2, 4, 4, 3)less than "S".  So below "S" we write 5 and divide it by 2! (two repeating alphabets)

Numerical value on "C" is 1
LHS of letter "C" there is no letter with less numerical value than "C".  So below "C" write 0 and divide it by 2! (two repeating alphabets)

Numerical value on "H" is 2
LHS of letter "H" there is no letter with less numerical value than "H".  So below "H" write 0 and divide it by 2! (two repeating alphabets)

Numerical value on "O" is 4
LHS of letter "O" there is one letter with less numerical value than "O".  So below letters "O" write 1 and divide it by 2! (two repeating alphabets)

After "O" there is again a letter "O"
Numerical value on "O" is 4
LHS of letter "O" there is one letter with less numerical value than "O".  So below letters "O" write 1 and divide it by 2! (two repeating alphabets)

Numerical value on "L" is 3
LHS of letter "L" there is no letter with less numerical value than "L".  So below "L" write 0
7) Now multiply the numbers in the third row with the numbers in the fourth and all these

8) Rank of the given given word is = Above total +1
= 302 + 1 = 303

With the help of above step we find the rank of the word SCHOOL as given below

## Find the rank of  " SCHOOL"

$\begin{matrix} 5&1 &2 & 4 & 4 & 3 \\S&C &H & O & O & L \\ 5!& 4! & 3! & 2! & 1! & 0! \\ \frac{5}{2!}&\frac{0}{2!} & \frac{0}{2!} &\frac{1}{2!} &1 & 0 \\\end{matrix}$

Now Calculate the following

$5!\times \frac{5}{2!}+4!\times 0+3!\times 0+2!\times \frac{1}{2}+1!\times 1+0$

$120\times \frac{5}{2}+0+0+2\times \frac{1}{2}+1$

= 60 ✖ 5 + 1+1

=300 + 2 = 302

Rank of the word SCHOOL is = 302+1 = 303

Some more examples are as follows

## Find the rank of  " COMPUTER"

Here total letters in the word computer is 8 with no repetition. So in 3rd row no need to divide with any number.
In the second row write the given number with capital letters.
In first row give numerical value alphabetically.
In the third row write the numbers 7!, 6! .......0! below the alphabets of the given word.
In the fourth row write the number of alphabets in the RHS of each alphabet which have less numerical value.
Then multiply the numbers in the 3rd row and 4th row.

$\begin{matrix}1& 4 & 3 & 5 & 8 & 7 & 2 & 6 \\ C & O & M & P & U & T & E & R \\ 7!& 6! &5! & 4! & 3! & 2! & 1! & 0! \\ 0& 2 & 1 & 1 & 3 & 2 & 0 & 0 \\\end{matrix}$

Now Calculate the following

= 7! ✖ 0 + 6! ✖ 2 + 5! ✖ 1 + 4! ✖ 1 + 3! ✖ 3 + 2! ✖ 2+ 1! ✖ 0 + 0! ✖ 0
= 0 + 720 ✖ 2 + 120 + 24 + 18 + 4 + 0 + 0
= 1440 + 166 = 1606
Rank of COMPUTER" is  1606 + 1 = 1607

## Find the rank of  " ASSEMBLY"

Here total letters in the word ASSEMBLY is 8 with 2 S.

$\begin{matrix}1& 6 & 6 & 3 & 5 & 2 & 4 & 7 \\ A & S & S & E & M & B & E & Y \\ 7!& 6! &5! & 4! & 3! & 2! & 1! & 0! \\ 0& \frac{4}{2!} & 4 & 1 & 2 & 0 & 0 & 0 \\\end{matrix}$

Now Calculate the following

= 7! ✖ 0 + 6! ✖ $\frac{4}{2!}$  + 5!  ✖ 4 + 4! ✖ 1 + 3! ✖ 2 + 0 + 0 + 0
= 0+ 720 ✖ 2 + 120 ✖ 4 + 24 + 12
= 1440 + 480 +36
= 1440 + 516 = 1956
Rank of ASSEMBLY" is  1956 + 1 = 1957

## Find the rank of  " EQUATION"

Here total letters in the word EQUATION is 8 with no repetition. So in 3rd row no need to divide with any number.

$\begin{matrix}2 & 6 & 8 & 1 & 7 & 3 & 5 & 4 \\ E& Q & U & A & T & I & O & N \\ 7!& 6! & 5! & 4! & 3! & 2! & 1! & 0! \\ 1& 4 & 5 & 0 & 3 & 0 & 1 & 0 \\\end{matrix}$

Now Calculate the following

= 7! ✕ 1 + 6! ✕ 4 + 5! ✕ 5 + 4! ✕ 0 + 3! ✕ 3 + 2! ✕ 0 + 1! ✕ 1 + 0! ✕ 0
= 5040 +720 ✕ 4 + 120 ✕ 5 + 0 + 6 ✕3 + 0 +1+0
= 5040+ 2880+600+18+1
= 8539
Rank of EQUATION" is  8539 + 1 = 8540

## Find the rank of  " UMBRELLA"

Here total letters in the word UMBRELLA is 8 with 2 L.

$\begin{matrix}7 & 5 & 2 & 6 & 3 & 4 & 4 & 1 \\ U& M & B & R & E & L & L & A \\ 7!& 6! & 5! & 4! & 3! & 2! & 1! & 0! \\ \frac{7}{2!}& \frac{5}{2!} & \frac{1}{2!} & \frac{4}{2!} & \frac{1}{2!} & \frac{1}{2!} & 1 & 0 \\\end{matrix}$

Now Calculate the following
7! ✕ $\frac{7}{2!}$ + 6! ✕$\frac{5}{2!}$ + 5!  ✕ $\frac{1}{2!}$ + 4! ✕$\frac{4}{2!}$ +3! ✕ $\frac{1}{2!}$+ 2! ✕$\frac{1}{2!}$ + 1! ✕1 + 0

17640 + 1800 +60 + 48 + 3 + 1 + 1
19553

Rank of UMBRELLA" is  19553 + 1 = 19554

Similarly find the rank of the following words

Ans: 135500