Counting words with specific requirements How many words of length exactly 6 can be created from letters A,B,C,D,E,F, so that each word has not more than one A, not more than one B, not less than one C and not less than one D?
I've tried using simple binomial coefficient, but it is clear that using this method one would count the same word multiple times.

We can break this up into 4 cases:
1) If the word has 0 A's and 0 B's, then there are $4^6$ possible words, and subtracting the words with no C's and no D's gives $4^6-2(3^6)+2^6=2702$ possibilities.
2) If the word has 1 A and 1 B, then there are 6 choices for the A, 5 choices for the B, and then $[4^4-2(3^4)+2^4]$ ways to complete the word so it contains at least one C and at least one D; so there are $6(5)[4^4-2(3^4)+2^4]=3300$ possibilities.
3) If the word contains 1 A and 0 B's, there are 6 choices for the A, and then
$[4^5-2(3^5)+2^5]$ ways to complete the word; so there are $6[4^5-2(3^5)+2^5]=3420$ possibilities.
4) If the word contains 1 B and 0 A's, then there are $6[4^5-2(3^5)+2^5]=3420$ possibilities as in case 3.
Thus there are $12,842$ possible words.

An alternate approach would be to use exponential generating functions:
Since $g_{e}(x)=(1+x)^{2}(e^{x}-1)^2e^{2x}=(1+2x+x^2)(e^{4x}-2e^{3x}+e^{2x})$, 
the coefficient of $x^6$ in $g_{e}(x)$ is given by
$\displaystyle\frac{a_{6}}{6!}=\frac{4^6}{6!}-2\cdot\frac{3^6}{6!}+\frac{2^6}{6!}+2\cdot\frac{4^5}{5!}-4\cdot\frac{3^5}{5!}+2\cdot\frac{2^5}{5!}+\frac{4^4}{4!}-2\cdot\frac{3^4}{4!}+\frac{2^4}{4!}$; 
so the number of possible words is given by
$a_{6}=4^6-2(3^6)+2^6+2(6)(4^5)-6(4)(3^5)+6(2)(2^5)+6(5)(4^4)-6(5)(2)(3^4)+6(5)(2^4)$.
