senior wrote:
I'm almost 16 years out of school, I found this one stumped me for a bit.
My preferred approach is to take all possibilities into account, and then backing out the restrictions.
Let's try a different problem with your approach. There are five people: A, B, C, D, E. Need to chose 3 for a committee. A and B cannot be chosen together. B and C cannot be chosen together. How many options?
Your approach: total 10 options, 5c3.
Now, assume the wrong scenario where A and B are chosen together. There are three such scenarios. (A and B are chosen, just need one more person.) So we have to subtract the three wrong options. Similarly, there are three wrong scenarios where B and C are chosen together.
This gives us 10-3-3=4 as the answer.
Yet this answer is wrong. There are five possibilities: ACD, ACE, ADE, BDE, CDE.
Do you see your mistake?
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Sergey Orshanskiy, Ph.D.
I tutor in NYC: http://www.wyzant.com/Tutors/NY/New-York/7948121/#ref=1RKFOZ