46=13+33, so any score with a 46 point dart can also be achieved by substituting a 13 point dart plus a 33 point dart.  Similar arguments can be made with 44=11+33, 42=11+31, and 33=11+11+11.  Then for any set of darts with some combination of all 7 possible scores can be mapped to a corresponding set with only the 11, 13, and 31 point darts.  (This will most likely be a many-to-one relation.)

Also 31+13=11+11+11+11, then this breaks down further that any solution with both 31 point darts and 13 point darts can be reduced to having at most some 31 or some 13, by substituting 11 point darts.  Then we have two cases: any score can be reduced to just 11 and 13 point darts OR just 11 and 31 point darts.

So then to make the goal of 100 points we then have "primitive" sets of darts with just 11 and 13 point darts or just 11 and 31 point darts.

Each of 11 and 31 are congruent to 1 mod 10.  This implies we need 10 darts to make a total score that is a multiple of 10.  But the smallest way to do that is 11*10=110, which is larger than 100 so no solutions in this case.

Darts of 11 and 13 can total 100 by 13*6+11*2=100.  This is exactly 8 darts.  Changing any 11 to 13 or 13 to 11 will result in a total not equaling 100.  7 darts is not enough as 7*13=91; 9 darts is too much since 9*11=99, and changing any 11 to a 13 dart will make the total 101 or larger.  Therefore the only "primitive" solution is to make 100 using 13*6+11*2=100.

Now to look at our substitutions: 31+13=11+11+11+11, 46=13+33, 44=11+33, 42=11+31, and 33=11+11+11.  We have only two 11's available so we cannot get a 31 or 33 to make a better solution and since we won't have 31 or 33 then 42, 44, or 46 also cannot appear in a solution.  Then the ONLY way to score exactly 100 points is by using eight darts: six darts scoring 13 points and 2 darts scoring 11 points.