How many ways can $4$ clients fill $7$ boxes if there may not be more than one client per box?

Question

I have a statement that says:

In a supermarket there are $7$ boxes available to use. If there are $4$ clients who want to pay for their purchases, in how many ways can the available boxes be filled, if there may not be more than one client per box?

My development was:

Each client have $7$ available boxes, are $4$ clients, so $7 \cdot 7 \cdot 7 \cdot 7 = 7^4$, also in each box, there may be $1$ or $0$ clients, then that would be $50$ of probability that there is a client or not ($1$ or $0$), then multiplied by $\frac{1}{2}$

So, finally the result is: $\frac{7^4}{2} = \frac{2401}{2}$, but my problem is that the correct result must be $840$, so, ¿Why is my result wrong?

Answer 1

The first client will be given one of the $7$ boxes. After that, the next client will be given one of the $6$ remaining boxes, next $5$, next $4$.

So there are $7\cdot 6\cdot 5\cdot 4=\frac{7!}{3!}=840$ ways.

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What you did wrong was giving the next clients also $7$ choices of boxes, which means the same box may be distributed among more clients, which was assumed not to be allowed. Also there are no probabilities needed in this exercise.

Answer 2

The first client arrives, and has seven possible boxes to choose from. Immediately after he has chosen his box, the second customer comes along. He sees that one box is taken, so he chooses from among the remaining six. In total, that's $7\cdot6=42$ different ways to distribute the first two customers.

The remaining two customers follow similar arguments to arrive at a total of $840$.