10. How many four-digit numbers can be formed using 3 ones, 2 twos, and 5 threes?

Using the ten digits consisting of 3 ones, 2 twos, and 5 threes, how many four-digit numbers can be formed?
10. According to the problem, the generating function $G_{e}(t)=\left(1+\frac{t}{1!}+\frac{t^{2}}{2!}+\frac{t^{3}}{3!}\right)\left(1+\frac{t}{1!}+\frac{t^{2}}{2!}\right)\left(1+\frac{t}{1!}+\frac{t^{2}}{2!}+\frac{t^{3}}{3!}+\frac{t^{4}}{4!}+\frac{t^{5}}{5!}\right)$ $=\left(1+2 t+2 t^{2}+\frac{7}{6} t^{3}+\frac{5}{12} t^{4}+\frac{1}{12} t^{5}\right)\left(1+\frac{t}{1!}+\frac{t^{2}}{2!}+\frac{t^{3}}{3!}+\frac{t^{4}}{4!}+\frac{t^{5}}{5!}\right)=\cdots+\left(\frac{5}{12}+\frac{7}{6}+1+\frac{1}{3}+\right.$ $\left.\frac{1}{24}\right) t^{4}+\cdots, \frac{t^{4}}{4!}$ The coefficient of $\frac{t^{4}}{4!}$ is 71. That is, 71 four-digit numbers can be formed.