Number of permutations of a string 
Calculate the number of permutations that is created by picking 4 letters from the string 'HARD CORD' (2 duplicates)

I know that if the string had 8 distinct letters, I can use $_8$P$_4$ to get the required answer. But how do things change when there are duplicates in the given string like the one given above?
 A: I think the easiest way to do this is to count the words with all letters different, then the other cases.
The number of words with all letters different is ${}^6\mathrm P_4=360$, since there are $6$ different letters to choose from.
The number of words with two Rs and other letters all different is ${}^4\mathrm C_2\times {}^5\mathrm P_2=120$, since there are ${}^4\mathrm C_2$ ways to choose the positions of the Rs and then ${}^5\mathrm P_2$ ways to pick two different letters from the remaining five, in order. Similarly there are $120$ words with two Ds and all other letters different.
Finally, the number of words with two Rs and two Ds is ${}^4\mathrm C_2=6$: once we choose two places for the Rs, we know where the Ds have to go.
This gives a total of $606$ words.
A: Q:"Calculate the number of permutations that is created by picking 4 letters from the string 'HARD CORD'" 
Okey lets analyse this one. First of all "permutation" by definition is a word, in which each letter appears exactly once. 
We have an Alphabet $X={A,C,D,R_1,R_2,H}$ I have to "pick" four letters from this Alphabet, and the result has to be permutation. 
Okey what could happen:
You could get 0R,1R,2R and 0D,1D,2D If (1D0R),(1R0D) and (1R1D) you are okey. 
So we have 3 possible disjunctive outcomes: 
1) (1D0R) and 3 random from { A,C,H } 
2) (1R0D) and 3 random from { A,C,H } 
2) (1R1D) and 2 random from {A,B, H  }
Lets count how many ways we could get 1). 
Get one D, Choose 3 out of 3, and count all permutation $1*C_3^3*4!$ 
Now the second part 2) Its the same as 1) 
Now the third part 
Simply choose 2 out of 3, and count all permutations. $1*C_3^2*4!$
Add 1)+2)+3) and you get the result.
The Formulation of the Questions is critical in combinatorics.Does someone sees a mistake in my argumentation? Edit, oversaw one D, in the first answer,....was blind!
