# Counting problem (students assigned to a tutor)

Four new students have to be assigned to a tutor. There are seven possible tutors, and none of them will accept more than one new student. In how many ways can the assignment be carried out?

The Answer is 840 but I don't understand how ? Isn't just $\binom{7}{3}$ ?

Remark that $840=24\times 35=4!\times \begin{pmatrix}7\\4\end{pmatrix}$. The exercise probably states somewhere that all students are different (this is where the $4!$ comes from).
Also $\begin{pmatrix}7\\4\end{pmatrix}=35$ should be there because you need to select $4$ tutors out of the seven tutors and then simply assign them the students. In other words if $NS_1,\dots NS_4$ are the new students and $T_1,\dots T_7$ are the tutors, "$NS_1$ assigned to $T_1$, $NS_2$ assigned to $T_2$, $NS_3$ assigned to $T_3$ and $NS_4$ assigned to $T_4$" is different from "$NS_4$ assigned to $T_1$, $NS_3$ assigned to $T_2$, $NS_2$ assigned to $T_3$ and $NS_1$ assigned to $T_4$".
Let the students be assigned one by one. For the first student there are $7$ options. Then for the second there are $6$ options left. Et cetera. $$7\times6\times5\times4=840$$