What is the rank of COCHIN Is there any shortcut method for finding the rank of the word COCHIN?
I mean is there any shortcut method for finding the rank of a word having repeated letters.
For example there is a shortcut method for finding the rank of the word having non repeated letters like PARLOUS as shown below

 A: Let's begin by counting distinguishable arrangements of the letters of the word COCHIN.  It contains six letters, so we have six positions to fill.  We fill two of those six positions with a C in $\binom{6}{2}$ ways, then arrange the O, H, I, and N in the remaining four positions in $4!$ ways.  Hence, there are 
$$\binom{6}{2} \cdot 4! = \frac{6!}{2!4!} \cdot 4! = \frac{6!}{2!} = 360$$
distinguishable arrangements of COCHIN.  
Method 1:  If we arrange the letters of COCHIN in alphabetical order, we obtain C, C, H, I, N, O.  By symmetry, $2/6 = 1/3$ of these arrangements will have a C in the first position.  Once we place a C in the first position, we have five positions to fill with the letters C, H, I, N, O.  By symmetry, $4/5$ of the arrangements that have a C in the first position will have a letter in the second position that precedes O alphabetically. Of the arrangements in which C is in the first position and O is in the second position, COCHIN is the first to appear in a lexicographical ordering of its letters.  Hence, COCHIN appears in position 
$$1 + \frac{4}{5} \cdot \frac{1}{3} \cdot 360 = 97$$
in a lexicographical ordering of its letters.  
Method 2:  To adapt your method, we make a table.
$$
\begin{array}{l l}
\text{chosen letter} & \text{available letters}\\
C & C, C, H, I, N, O\\
O & C, H, I, N, O\\
C & C, H, I, N\\
H & H, I, N\\
I & I, N\\
N & N
\end{array}
$$
If we place C, O, C, H, I, N in that order in the six positions for the letters, the first C is the first of the six letters, O is the fifth of the five remaining letters, the second C is the first of the four remaining letters, H is the first of the three remaining letters, I is the first of the two remaining letters, and N is the only remaining letter, we obtain
$$0 \cdot 5! + 4 \cdot 4! + 0 \cdot 3! + 0 \cdot 2! + 0 \cdot 1! + 0! = 97$$
for the position of COCHIN in a lexicographical ordering of its letters, in agreement with the symmetry argument given above.  
In your method, the first number in each product represents the number of available letters that precede the chosen letter, while the second number represents the number of ways of arranging the remaining letters.  Thus, we were somewhat fortunate that COCHIN begins with a C since that meant that once we placed a C in the first position, the remaining letters were all distinct, which made it easier to count the permutations of the remaining letters.  We were similarly fortunate in using the symmetry argument because there are $4!$ arrangements that begin with CC, CH, CI, and CN.
Method 3:  We must have a C in the first position.  Of the words that begin with CO, COCHIN is the first to appear in the lexicographical ordering of its letters.  Thus, the words that precede COCHIN in a lexicographical ordering of its letters begin with CC, CH, CI, or CN.  
$$
\begin{array}{c c}
\text{initial letters} & \text{remaining letters}\\
CC & H, I, N, O\\
CH & C, I, N, O\\
CI & C, H, N, O\\
CN & C, H, I, N\\
\end{array}
$$
For each of the four ways of choosing the first two letters of the words that precede COCHIN in the lexicographical ordering of its letters, there are $4!$ of arranging the remaining letters.  Hence, there are $4 \cdot 4!$ words that precede COCHIN in a lexicographical ordering of its letters.  Hence, its ranking is 
$$4 \cdot 4! + 1 = 97$$
Note:  Of these methods, I recommend that you use the third one for the reason true blue anil illustrates by finding the ranking of SUCCESS in the lexicographic ordering of its letters.
A: $\underline{Revised\; and\; improved\; answer}$
I take it that your question is for repeating letters in general, and the warning is: The shortcut method above without proper modification doesn't work for words with repeated letters, although some sites on the net suggest otherwise
[ The reason for this will become clear as we proceed. ]
Let us illustrate taking the word SUCCESS
Letters fixed from the left will be enclosed in $[...]$, the $ ...$ for computing permutaions of the remaining letters.
Starting turn by turn with the lowest ranking letter(s), we get
$[C......]: 6!/3! = 120$
$[E......]: 6!/3!2! = 60$
$[SC.....]: 5!/2! = 60$
$[SE.....]: 5!/2!2! = 30$
$[SS.....]: 5!/2! = 60$
$[SU.....]: 1 [$ The remaining letters, $CCESS$ are already in lexicographic order.]
Why the "original" shortcut method doesn't work here is because, for example, permutations starting with $SC$ aren't the same as permutations starting with $SE$
However, the method, properly modified to take into account differences in permutations due to  repeated letters, does work, as explained, e.g. here
The basic idea is to count how many words are formed before the $S$ becomes the first word , then how many words from the remaining letters are formed before the required $U$ turns up, and so on.
A: To answer to the question asked, there is is a shortcut method available. See this online tool  which finds rank of any word using shortcut method. This method is working fine for words with both repeated and non-repeated letters see my question where I had a discussion about this tool.
(acutally the tool uses animation for explaining. however, posting the screenshots for anyone who is interested in shortcut method)

For SUCCESS,  rank is $331$ (output of the tool is given below)



For Cochin, rank is $97$ (output of the tool is given below)


