How many ways can a tennis player be scheduled to play $4$ matches in $8$ days, with at most one match per day? Should I use combinations or permutations?

How many ways can a tennis player be scheduled to play $4$ matches in $8$ days, with at most one match per day?

Currently, I work mostly in combinatorics to catch up on. My guess is that we should use $V_{8}^{4}=8.7.6.5=1,680$. The correct answer according to my book is $70$ that is equal to $C_8^4=\dfrac{V_{8}^{4}}{P_4}=\dfrac{1680}{4.3.2}=70$. In my opinion the order of games matters.

The question asks for the number of ways it is possible to select the days he plays a tennis match.  There are only 
$$\binom{8}{4} = \frac{8!}{4!4!} = 70$$
ways to select four of the eight days for him to play a tennis match.
It does not say anything about his opponents.  In particular, it does not specify that he is playing against different opponents.  Your answer 
$$P(8, 4) = \frac{8!}{(8 - 4)!} = 8 \cdot 7 \cdot 6 \cdot 5 = 1680$$
counts the number of ways the tennis player could play four matches against different opponents in eight days if he does not play more than one match per day. Notice that if he played against his practice partner each time, only the days on which he plays a match would matter. 
