Number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed by using the letters of the word DIFFERENTIATION 
Find the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed by using the letters of the word DIFFERENTIATION.

Vowels are IIIEEAO. Consonants are DFFNNTTR.Vowels can be selected in $\binom{4}{3}$ ways and consonants can be selected in $\binom{5}{4}$ ways and total ways are $\binom{4}{3}\binom{5}{4}7!=100800$ ways, but the answer is 532770. Where did I go wrong?

Another approach worth mentioning is that of exponential generating functions (egfs). Let $x$ enumerate vowels, $y$ enumerate consonants and $z$ enumerate total letters. Then the egf factors for vowels break down as follows
III
$$1+xz+\frac{1}{2!}(xz)^2+\frac{1}{3!}(xz)^3$$
EE
$$1+xz+\frac{1}{2!}(xz)^2$$
A, O
$$1+xz$$
and consonants:
FF, NN, TT
$$1+yz+\frac{1}{2!}(yz)^2$$
D, R
$$1+yz$$
The product of these is the egf for words 
$$\begin{multline}f(x,y,z)=\left(1+xz+\frac{1}{2!}(xz)^2+\frac{1}{3!}(xz)^3\right)\times\\\left(1+xz+\frac{1}{2!}(xz)^2\right)\left(1+\vphantom{\frac{1}{1}}xz\right)^2\left(1+yz+\frac{1}{2!}(yz)^2\right)^3\times\\\left(1+\vphantom{\frac{1}{1}}yz\right)^2\end{multline}$$
and we want the $x^3y^4z^7/7!$ coefficient of $f(x,y,z)$. Using the sage input
y,z=var('y','z')
show(factorial(7)*expand((1+z*x+(z*x)^2/2+(z*x)^3/6)*(1+z*x+(z*x)^2/2)*(1+z*x)^2*(1+z*y+(z*y)^2/2)^3*(1+z*y)^2).coefficient(z^7).coefficient(x^3).coefficient(y^4))

gives our answer:

$$532\,770\tag{Answer}$$

