Shooting Mania

The number of ways he can score 30 points is the same as the number of ways he can be 5 points short of a perfect score.

The number of ways to distribute 5 points across 7 shots is ${{5+6}\choose{5}}$ , using a "stars and bars" calculation. Since there are no restrictions on the arrangements (i.e. it is impossible for him to miss more than 5 points in a single shot, because he only missed 5 points total), the final answer is:

${11\choose{5}}=\boxed{462}$

(11,6)=462

4 4 4 4 4 5 5 30 21

3 4 4 4 5 5 5 30 140

3 3 4 5 5 5 5 30 105

2 4 4 5 5 5 5 30 105

2 3 5 5 5 5 5 30 42

1 4 5 5 5 5 5 30 42

0 5 5 5 5 5 5 30 7

                            462

Possible arrangements of numbers are as follows: $1) \quad 0 + 5 + 5 + 5 + 5 + 5 + 5 = 30 \qquad 7 \qquad \qquad \qquad$ $2) \quad 4 + 4 + 4 + 4 + 4 + 5 + 5 = 30 \qquad {7 \choose 2} = 21 \qquad$ $3) \quad 3 + 3 + 4 + 5 + 5 + 5 + 5 = 30 \qquad {7 \choose 4} \times 3 = 105$ $4) \quad 2 + 4 + 4 + 5 + 5 + 5 + 5 = 30 \qquad {7 \choose 4} \times 3 = 105$ $5) \quad 3 + 4 + 4 + 4 + 5 + 5 + 5 = 30 \qquad {7 \choose 3} \times 4 = 140$ $6) \quad 2 + 3 + 5 + 5 + 5 + 5 + 5 = 30 \qquad {7 \choose 5} \times 2 = 42$ $7) \quad 1 + 4 + 5 + 5 + 5 + 5 + 5 = 30 \qquad {7 \choose 5} \times 2 = 42$

The result is: $7 + 21 + 105 + 105 + 140 + 42 + 42 = \boxed{462}$