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Continuing on with the GRE practice questions I have, I'm confused about how to solve the following types of probability questions.

  1. Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat?

    There are 120 ways they can sit together in a row of 5 seats because 5! = 120. I'm confused as to how you would mathematically describe a situation where Martha is sitting only in the "middle" seat.

  2. How many 3-digit positive integers are odd and do not contain the digit 5 ?

    If I am correct (probably not), there are 999–99=900 3-digit positive integers (this means there are 450 odd 3-digit positive integers, I think). Of these, 100/900 will be a 5 (i.e. 5xx), 10/100 10s will be a 5 (i.e. x5x), and 1/10 1s digits will be a 5 (i.e. xx5). At this point I'm lost in numbers and it has taken me much longer than the minute I will be given to solve it. Am I missing some sort of heuristic to solving these?

Thanks for you help.

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These are actually combinatorics problems rather than probability problems. This sort of combinatorics is needed for doing probability problems, so they may appear in sections of texts and courses that are labeled "probability". –  Michael Hardy Jul 29 '11 at 22:09

1 Answer 1

up vote 4 down vote accepted
  1. Fix the middle seat for Martha. Now you have to position the other 4 people in 4 seats: $4!=24$.
  2. Build the number by the digits: you have $8$ options for the hundreds (all but 0 and 5), $9$ options for the tens (all but 5) and $4$ options for the units (only one of $1,3,7,9$). A total of $8\cdot 9\cdot 4$ numbers.
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That's exactly what I was looking for! Awesome, thanks! –  stoicfury Jul 29 '11 at 21:50

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