Let N(s,d) be the number of ways to achieve a score of s using exactly d darts. In each throw we can get a score of either 0, 1, 2, or 3 so if we scored
0 then N(s,d) = N(s,d-1) (still need to get s but with one less dart)
1 then N(s,d) = N(s-1,d-1) (scored one, so need to score s-1 with one less dart)
2 then N(s,d) = N(s-2,d-1) (scored 2 so need s-2 with one less dart)
3 then N(s,d) = N(s-3,d-1) (...)
But we could score any of these, so
Where N(0,0)=1, N(s<=0,d)=0, and N(s,d<=0)=0
I'm sure that there is an exact formula for N, probably with lots of factorials, but in much less than the time it would take me to figure it out, I could write a computer program to do it for me. So I'll do that.
Here is the output of the program. The rows are the score and the colums are the number of darts.
0 1 2 3 4 5 6
0| 1 1 1 1 1 1 1
1| 0 1 2 3 4 5 6
2| 0 1 3 6 10 15 21
3| 0 1 4 10 20 35 56
4| 0 0 3 12 31 65 120
5| 0 0 2 12 40 101 216
6| 0 0 1 10 44 135 336
7| 0 0 0 6 40 155 456
8| 0 0 0 3 31 155 546
9| 0 0 0 1 20 135 580
10| 0 0 0 0 10 101 546
11| 0 0 0 0 4 65 456
12| 0 0 0 0 1 35 336
13| 0 0 0 0 0 15 216
14| 0 0 0 0 0 5 120
15| 0 0 0 0 0 1 56
16| 0 0 0 0 0 0 21
17| 0 0 0 0 0 0 6
18| 0 0 0 0 0 0 1
19| 0 0 0 0 0 0 0
Using this we can see that the number of ways to score 15 with 6 dart is 56, and the number of ways to score 16 is 21. So 56+21=77 which is your answer.
Hope that helps,
Kev