Solving it using stars and bars method,
let x1,x2,x3 and x4 be their scores when rolling.
we need x1+x2+x3+x4 = 17 ($1\leq Xi \leq 6$)
since they have minimum value 1, remaining 13 have to be scored as,
x1'+x2'+x3'+x4' = 13 (Xi' = Xi-1)
N1 = number of ways = c(13+4-1 ,13) = c(16,3) = 560 .
Now we have to subtract the number of ways in which atleast one of Xi $\geq$ 7 .
assuming x1 as 7 , since it is already having minimum 1, we will assume x1'=6 . If we split the remaining among 4 now, any split will be invalid as x1 is already invalid.
x1'+x2'+x4'+x4' = 7 (since x1' is already a 6)
Number of ways of splitting 7 among all 4 = C(7+4-1,7) = C(10,3) = 120
similarly same case can come for x2', x3' and x4' . which will make it a total of ,
N2 = 4*120 = 480.
Now we have some duplicate cases , such as,
1.when we fix x1'=6 and we split and if we give x2'=6
2. when we fix x2'=6 and we split and give x1'=6.
here case 1 and case2 are same. i.e., we have to subtract all cases from N2 where two of the Xi get value more than 6. which can be done in ,
x1'+x2'+x3'+x4' = 1 (since two have now value 6)
ways = C(1+4-1,1) = C(4,1) = 4.
we can choose C(4,2) such duplicate pairs gives a total of
N3= C(4,2)*4 = 24.
Total number of ways = N1-(N2-N3) = 560 - (480-24) = 104 and hence the answer.