1. Given that there are 6 parallel network lines between locations $A$ and $B$, and their maximum information capacities are $1,1,2,2,3,4$. Now, if we randomly select three network lines such that the sum of their maximum information capacities is greater than or equal to 6, the number of ways to do this is ( ) ways.
(A) 13
(B) 14
(C) 15
(D) 16
-,1.C.
As shown in Figure 6, according to the problem, at least one of the fifth and sixth network cables must be used.

If the sixth network cable is chosen, any two of the first five can be selected, which gives a total of 10 ways;

If the fifth network cable is chosen, then selecting the third and fourth (3, 2, 2) has 1 way; or selecting one from the third and fourth and one from the first and second (3, 2, 1) has $2 \times 2=4$ ways, in this case, there are $1+4=5$ ways.
In summary, there are $10+5=15$ ways.