Here is the question I worked on and got the correct answer but not sure why this works:
Problem #34 in the 5lb book of practice problems for the GRE by Manhattan prep:
Jan and 5 other children are in a classroom. The principal will choose two children at random, what is the probability that Jan will be chosen?
Probability of A = Jan chosen first = 1/6
Probability of B = Jan not chosen x Jan Chosen = 5/6 x 1/5 = 1/6
Mutually exclusive events can be added therefore the Probability of A or B = 1/6 +1/6 = 1/3
The book went into a counting principle instead - wondering if this shortcut will work for other problems?
It’s a standard way to do the problem, so i wouldn’t call it a shortcut.
Alternatively, you could’ve also done:
\frac{5}{6\operatorname{C}_2}
to stay more in line with the “counting principle” topic depending on what is covered in the book.
Both are correct, Notice that order is not mattering in the question above. And in your method you are making that sure by taking both cases. Counting way would look like this,
We need two people, 1 Jan is fixed and other can be anyone. Hence total ways to do that is 1*5C1 = Favourable cases
We need two people at random, hence total ways would be 6C2 = Total Cases
Probability = Favourable Cases/ Total Cases = 5/6C2 = 1/3
Note: With your method it’s easy to do until only 1 person is fixed, But will become tricky if 2 people are fixed. Because then you will have to take lot more cases. And there counting method would be easiest.
Hope this helps as explanation. 
Thank you both - these are very helpful!
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