4) A family consisting of two parents and two young children wants to cross a river. Their boat can carry at most two young children or one adult. Counting both the crossings in one direction and those in the other, what is the minimum number of crossings the boat must make? (Obviously, the boat cannot cross the river without being manned).
(A) 5
(B) 6
$\begin{array}{ll}\text { (C) } 7 & \text { (D) } 8\end{array}$
(E) 9.
4) The answer is (E). Let's denote the parents as $G^{\prime}$ and $G^{\prime \prime}$, and the children as $F^{\prime}$ and $F^{\prime \prime}$. We will show that 9 crossings are sufficient. The strategy is: (1) $F^{\prime}$ and $F^{\prime \prime}$ go; (2) $F^{\prime}$ returns; (3) $G^{\prime}$ goes; (4) $F^{\prime \prime}$ returns; (5) $F^{\prime}$ and $F^{\prime \prime}$ go; (6) $F^{\prime}$ returns; (7) $G^{\prime \prime}$ goes; (8) $F^{\prime \prime}$ returns; (9) $F^{\prime}$ and $F^{\prime \prime}$ go. To see that fewer crossings are not possible, observe that (apart from the possible exchange between the parents or between the children when it is possible) the strategy is mandatory. Indeed, at each step, a different move would necessarily (since the boat cannot go by itself) result in round trips of the same people and increase the total number of crossings.

## Second Solution

Let's call $A$ the adults and $F$ the children. We will schematize the problem as follows: if, for example, one adult and two children are on the initial bank, one adult and the boat are on the final bank, then we indicate the situation as $(A F F, A B)$. To understand what the optimal sequence of trips is, we will rely on the following simple principle: if during a series of crossings the same situation occurs twice, the series is not optimal.

At the beginning, we are in the situation $(A A F F B, \cdot)$, and it is clear that the first trip must be undertaken by both children, otherwise whoever goes to the other bank will have to return with the boat, bringing us back to the starting point $(A A F F B, \cdot)$. So after the first trip, we will be in $(A A, F F B)$, and one of the children must bring the boat back (second trip), putting us in the situation $(A A F B, F)$. The third trip will be made by an adult (otherwise we would return to $(A A, F F B)$), and the fourth trip will be made by the child remaining on the arrival bank, putting us in the situation $(A F F B, A)$. The fifth trip cannot be undertaken by the adult or by one of the children alone, otherwise to bring the boat back, we would return either to the situation $(A F F B, A)$ or to $(A A F B, F)$. Therefore, after the fifth trip, we will be in $(A, A F F B)$, and continuing to reason in the same way, it is easily seen that the optimal transfer will necessarily pass through the situations $(A F B, A F)$, $(F, A A F B)$, $(F F B, A A)$, $(\cdot, A A F F B)$, for a total of four more crossings. Therefore, in total, the family will need at least 9 trips to transfer.