Are you smarter than me? 29

Let's first identify all sets (that is, disregarding order) of 7 elements among 5, 4, 3, 2 and 0 that sums up 30.

One easy way to think about it is to sum up six '5's, than five '5's, than four '5's, and so on, and see how you can use the other numbers to complete 30.

We'll see that the possible sets are:

1) {5, 5, 5, 5, 5, 5, 0}

2) {5, 5, 5, 5, 5, 3, 2}

3) {5, 5, 5, 5, 4, 4, 2}

4) {5, 5, 5, 5, 4, 3, 3}

5) {5, 5, 5, 4, 4, 4, 3}

6) {5, 5, 4, 4, 4, 4, 4}

But we have to take the order into account since we want to know how many ways, so we have to consider permutation with repetition giving by $\frac{n!}{r_{1}!r_{2}!...r_{m}!}$ where $n$ is the total number of elements in the permutation and $r$ is the number of times a single element appears in the permutation.

By disregarding elements that don't appear at all, and the ones that appear only once, since $0! = 1! = 1$ , set number 3, for example, would have $\frac{7!}{4!2!} = \frac{7.6.5}{2} = 105$ ways. Using the same logic for all 6 sets, and summing up all results, we would have $7 + 42 + 105 + 105 + 140 + 21 = 420$ ways of reaching 30 points.

Coefficient of $x^{30}$ of the generating function:

$g(x) = (1+x^{2}+x^{3}+x^{4}+x^{5})^{7}$

This can be solved using an organized casework:

5555550 can be arranged in 7 ways.

5555532 can be arranged in 42 ways.

5555442 can be arranged in 105 ways.

5555433 can be arranged in 105 ways.

5554443 can be arranged in 140 ways.

5544444 can be arranged in 21 ways.

7+42+105+105+140+21=420 :D