Number of ways to distribute $10$ different books to three children such that one child gets $3$ books, another gets $3$ books and the other gets $4$ books.
We have the number of ways to distribute as
$$\binom{10}{3}\times \binom{7}{3} \times \binom{4}{4}=\frac{10!}{3!3!4!}$$
But I have doubt whether we need to multiply above answer with $\frac{3!}{2!}$ since the distribution $3, 3,4$ can arrange themselves in $\frac{3!}{2!}$ ways.
Your worries are justified.
We have to select which child receives four books, which can be done in $3$ ways. We select four of the ten books for that child, three of the remaining six books for the older of the other two children, and give the remaining three books to the other child who receives three books. Thus, the number of ways the books may be distributed so that one child receives four books and two children each receive three books is $$\binom{3}{1}\binom{10}{4}\binom{6}{3}\binom{3}{3}$$ which agrees with the answer $$\frac{3!}{2!}\binom{10}{3}\binom{7}{3}\binom{4}{4}$$