What is the total number of study schedules for a seven day period for four subjects where each subject is allotted at least one day?

John Watson has provided you with a nice solution. We can obtain the same result using the Inclusion-Exclusion Principle.

For each of the seven days, the student has four choices of subject to study. Therefore, if there were no restrictions, she would have $4^7$ ways to choose her study schedule. From these, we must exclude those options in which she does not study each subject at least once.

She has $\binom{4}{k}$ ways to exclude $k$ of the four subjects from her study schedule and $(4 - k)^7$ ways to use the seven days to study the remaining $4 - k$ subjects. By the Inclusion-Exclusion Principle, the number of ways the student can arrange her schedule so that she studies each subject at least once over a period of seven days is $$\sum_{k = 0}^{4} (-1)^k\binom{4}{k}(4 - k)^7 = \binom{4}{0}4^7 - \binom{4}{1}3^7 + \binom{4}{2}2^7 - \binom{4}{3}1^7 + \binom{4}{4}0^7 = 8400$$

What is wrong with your attempt?

By designating particular days as the ones on which she studies each subject, you count distributions in which she studies a subject more than once multiple times.

Let's use John Watson's solution as a template.

You count schedules in which she studies one subject on four days and the other subjects on one day each four times, once for each way you could have designated one of those four days as the day she studies the subject she studies four times.

You count schedules in which she studies one subject on three days, a second subject on two days, and the other subjects on one day each six times, once for each of the three ways you could designate one of the three days as the day on which she studies the subject she studies three times and once for each of the two ways you could designate one of the two days as the day she studies the subject she studies on two days.

You count schedules in which she studies three subjects on two days each and one subject on one day eight times since each subject she studies on two days is counted twice, once for each of the two ways you could designate a particular day as the day on which she studies a subject she studies on two days.

Notice that $$\binom{7}{4} \cdot 4! \cdot 4^3 = \frac{7!}{3!} \cdot 4^3 = 4 \cdot \binom{4}{1} \binom{7}{4} \cdot 3! + 6 \cdot \binom{4}{1}\binom{7}{3}\binom{3}{1}\binom{4}{2}\binom{2}{1} + 8 \cdot \binom{4}{3}\binom{7}{2}\binom{5}{2}\binom{3}{2}$$

I'm not quite sure if this is correct. So we could have distributions of type $a)$ $$4,1,1,1$$ or b) $$3,2,1,1$$ or c) $$2,2,2,1$$ The number of distribution of type a) is $4\times {7\choose 4}\cdot 3\cdot 2 =840$

The number of distribution of type b) is $12\times {7\choose 3}\cdot {4\choose 2}\cdot{2\choose 1}=5040$

The number of distribution of type c) is $4\times {7\choose 2}\cdot {5\choose 2}\cdot {3\choose 2}=2520$

So we have $840+5040+2520 = 8400$ schedules.