Number of $ 6 $ Digit Numbers with Alphabet $ \left\{ 1, 2, 3, 4 \right\} $ with Each Digit of the Alphabet Appearing at Least Once Find the number of 6 digit numbers that can be made with the digits 1,2,3,4 if all the digits are to appear in the number at least once.
This is what I did -
I fixed four of the digits to be 1,2,3,4 .
Now remaining 2 places can be filled with 4 digits each. Number of 6 digit numbers if two places are filled with same digit are 4 * 6!/3! and if filled by different digits are 12 * 6!/(2!*2!). Therefore, total such numbers are 2880.
But the correct answer is 1560.
Any hint would be appreciated.

Really easy using generating functions. The generating function is:
$$G(x)=\left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right)^4$$
We need the coefficient of x^6 in that  and we make use of the fact that $x^4 = x^2 \cdot x^2$
$\left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right) \left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right)=x^2+x^3+\frac{7 x^4}{12}...$
We only need to calculate the first 3 terms of that because other terms do not contribute to $x^6$
$\left(x^2+x^3+\frac{7 x^4}{12}\right) \left(x^2+x^3+\frac{7 x^4}{12}\right)=\frac{7 x^6}{12}+x^6+\frac{7 x^6}{12}=\frac{13 x^6}{6}$
Now we have the required coefficient, we just multiply by $6!$
$$6! \cdot \frac{13}{6}= 1560 $$
That is the answer.
