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!
 A: 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.
A: The multiplicities of the individual letters NARGTSE are $2,3,2,2,2,1,1$. Then the number of $4$-letter words is the coefficient of $\frac{x^4}{4!}$ (that is $24$ times the coefficient of $x^4$) in the product
$$
  (1+\frac x1+\frac{x^2}2+\frac{x^3}6)(1+\frac x1+\frac{x^2}2)^4(1+\frac x1)^2
$$
(there is one factor $(1+\cdots+\frac{x^m}{m!})$ for each letter of mulitplicity$~m$). It is not too hard to work (the initial terms of) this product out by hand (though more work than what you showed in the question), but it is entirely mechanical and trivial using computer algebra. The full expansion is
$$
1+7x+\frac{47}2x^2+\frac{301}6x^3+76x^4+\frac{259}{3}x^5+\frac{303}4x^6
+\frac{625}{12}x^7+\frac{1351}{48}x^8+\frac{569}{48}x^9+\frac{365}{96}x^{10}
+\frac{85}{96}x^{11}+\frac{13}{96}x^{12}+\frac1{96}x^{13}
$$
and the term $76x^4$ gives $24\times76=1824$ which confirms the number you found. (It is rather surprising that this was an integer coefficient, and the other terms show that it is somewhat of an accident.) This full expansion will also easily give you the number of words of any other length that one can make from this stock of letters, for instance one can make $8!\times\frac{1351}{48}=1134840$ words of length$~8$.
See this answer and this one for detailed explanations of this "exponential generating polynomial" method for solving such scrabble-type counting problems.
