This solution by Knewton instructor is quite clear:

There are 20 students

The play has 20 characters

20 characters = 3 types

Each student is assigned to a unique character but character types (lead / supp / extra) can be repeated

hence from #1: We only know Supp type => insufficient by itself

from #2: We have extras but not supp /lead =.> insufficient by itself

Both together: we have lead =2, supp=6 and extras=12

=> Pr of J and J for one lead and one supp = 2/20 *6/19 + 2/20*6/19 = 6/95

Hence both together are sufficient

karthikpandian19 wrote:The twenty students in a class are to be randomly assigned parts in a play. There are twenty characters in the play, and each character is either a lead character, a supporting character, or an extra. Each student will be assigned exactly one part, and no two students will be assigned to play the same part. If Jannie and Jessica are two students in the class, what is the probability that one of the two is assigned to play a supporting character and the other is assigned to play a lead character?

Fourteen students will not be assigned to play supporting characters.
Eight students will not be assigned to play extras.

For this question, if we know how many supporting and how many leads there are we'd have sufficient data to determine what the probability is (since we know that total is 20).

With that in mind , let's look at the statements:

1. Fourteen students will not be assigned to play supporting characters.
This means that 20-14 = 6 students are supporting characters. We don't know how many leads there are. Insufficient.

2. Eight students will not be assigned to play extras.
This means that 20-8 = 12 play extras. We don't know how many supporting or leads are. Insufficient.

Together we have:
S = 6.
E = 12.
L must be 20-E-S = 2.

So, we have 2 leads and 6 supporting out of total 20. Sufficient data to determine probability.

C is the correct answer.