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Note: for the "modified" version of the Jane problem, the denominator should be

220 - 6C3 - 6C3 = 180

making the probability

45/180 = 1/4

The reason is that we have 6C3 ways of picking three juniors from six, and the same for picking three juniors from six. We subtract this from 220 since that's what we don't want (i.e, all juniors and all seniors) - this isn't 1 as Greg showed in the video.

You can also do this using permutations, which should result in the same answer. In that case you'll have the sample space (i.e, the number of ways of picking three students from 12 such that you have at least one junior and one senior) as 12×11×10-6×5×4-6×5×4, and the numerator would be 3!(1×6×5/2 + 1×5×6) - notice the division by 2. This is because we don't want duplicates - notice that [Jane, Senior 4, Senior 5] and [Jane, Senior 5, Senior 4] for instance is the same thing (which we already account for with the 3!).

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