The chance of the word to be of all different letters? A four letter word is written down by taking letters from the word KANYAKUMARI.What is the chance of the word to be of all different letters?

Since KANYAKUMARI has K,K,A,A,A,N,Y,U,M,R,I letters in which there are 2 K's and 3 A's.The chance that the four letter word to be of all different letters$=\frac{m}{n}$,where $m=\binom{8}{4}$,because there are 8 different letters K,A,N,Y,U,M,R,I.and $n=\binom{11}{4}$.So probability$=\frac{7}{33}$,but the correct answer is $\frac{840}{1109}$.I dont know where i have made the mistake.Please help me.
 A: You have to take permutations into account.
If we take the 8 distinct letters, there will be $\binom{8}{4}*4!= 1680$ distinct words
If we take all the 11 letters, the number of distinct words will be given by:
coefficient of $x^4$ in $4!(1+x)^6(1+x+x^2/2!)(1+x+x^2+x^3/3!)= 2218$
[Or else enumerate words that are  $3-1$ of a kind, $2-2$ of a kind, $2-1-1$ of a kind, etc]
$Pr = \dfrac{1680}{2218} = \dfrac{840}{1109}$
A: Favourable cases 
Here we have $8$ distinct letters , Then there will be $\displaystyle \binom{8}{4}\times 4! = 1680$
Here $\bf{KANYAKUMARI}$ contain $\bf{2K,3A,N,Y,U,M,R,I}$
Now Total no. of cases
$\bullet\; $ If all letters are different, Then we will take $4$ different letters
So total no. of arrangement $\displaystyle = \underbrace{\binom{8}{4}}_{\bf{select\; 4\; diff.\; letters}}\times \underbrace{4!}_{\bf{arrange\; these\; 4\; letters}}$
$\bullet\; $ If $2$ letters are same and other $2$ letters are different.
So we will take a $1$ pair of letter from $2$ pair letter and other $2$ from $7$ different letters
So tatal no. of ways $\displaystyle =\underbrace{\binom{2}{1}\times \binom{7}{3}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{4!}{2!}}_{\bf{arrangement\; of\; letters}}$
$\bullet\; $ Here $2$ same letters are of same kind and other $2$ are of same kind
So tatal no. of ways $\displaystyle \underbrace{\binom{2}{2}\times \binom{6}{2}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{4!}{2!\times 2!}}_{\bf{arrangement\; of\; letters}}$
$\bullet\; $ Here $3$ same letter are of same kind and other $1$ letter is different
So tatal no. of ways $\displaystyle = \underbrace{\binom{1}{1}\times \binom{7}{1}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{4!}{3!}}_{\bf{arrangement\; of\; letters}}$
So Total no. of ways $\displaystyle $ Sum of all above cases $\displaystyle =2218 $
So Required probability $\displaystyle = \frac{\bf{favourable\; cases}}{\bf{Total\; no.\; of \; ways}}$
A: Following Andre Nicolas's suggestion, we think of the letters as being on 11 "Scrabble" tiles, with all combinations of tiles equally likely, and with the tiles for the A's (and K's) painted with different colors.
Since each selection of 4 tiles can be arranged in $4!$ ways, we can disregard the ordering of the tiles, 
so we get the probability that all 4 letters are distinct is given by
$\hspace{.3 in}\displaystyle\frac{\binom{6}{4}+2\binom{6}{3}+3\binom{6}{3}+6\binom{6}{2}}{\binom{11}{4}}=\frac{205}{330}=\frac{41}{66}$
by considering the cases 
$\;\;\;\;$1) no A's, no K's $\;\;$2) 1K, no A $\;\;$ 3) 1 A, no K $\;\;$ 4) 1A, 1K
A: Let's look at it a bit differently.
Suppose you have a bag of balls. The color distribution is 
6 White
3 Red
2 Blue

What is the probability that if you grab 4 random balls, without replacement, that you have no more than one red ball, and no more than one blue ball?
Edit: This will only work if the original question is not allowing the reusing of letters. I am unsure from how it is currently phrased.
