Number of permutations of a word taking four at a time Take the letters $NNAAARRGGTTSE$. I have written my answer below to find out the number of permutations of four letters chosen from the given set of letters.
3 of the same kind and 1 other = $\left( \binom{6}{1} \cdot 4!\right)\big/3!$
2 of the same kind and another 2 from a similar kind = $\binom{5}{2}\cdot 4!\big/(2!2!)$
2 of the same kind and 2 different = $\binom{5}{1}\binom{6}{2}\cdot 4!/2!$
All four different = $7\choose4$*$4!$
Hence the answer I get is 1824. I hope someone could verify this for me, and if this method is too long please do suggest a short cut. Thanks!

Your way is perfect. Nothing wrong. 
(If you write this answer in an exam, I think you might need a bit more explanation. For example, an explanation about what $\binom{6}{1}$ means in the first case.)
Also, I don't know any better way than yours to solve this question.
