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Actuary P1 exam prep question

  1. Apr 2, 2014 #1
    1. Given a 6x5 array of unique elements, how many ways can three elements be selected so that no two are in the same row or column?

    2. Fundamental Counting Principle?

    3. This question was a sample question in the Actuary P1 exam prep I am reviewing. I thought it would be easy enough to use Counting Principle to determine the solution. So, the set of your first choice has 30 elements. Since this eliminates exactly one row and column, the set of your second choice only contains the remaining 4x5 array with 20 elements. And so on to get 12 elements for the third choice. Using FCP to determine the total number of combinations SHOULD therefore yield a solution of 7200. However, the provided solution for this question from the exam prep is different. What have I overlooked?
  2. jcsd
  3. Apr 2, 2014 #2

    Ray Vickson

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    I also get something very different from your 7200; I just chose the rows and columns independently. What answer was provided in the solution you saw?
  4. Apr 2, 2014 #3
    The given solution is 1200. What do you mean by selecting rows and columns independently? I don't really understand what is wrong with the way I approached this problem; could you maybe elaborate on what was wrong with my "solution."
  5. Apr 2, 2014 #4

    Ray Vickson

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    The provided solution is yours divided by 6 = 3!, and so recognizes that the three choices can be in any order; in other words, you are counting the same choice several times, because the concepts of "first", "second" and "third" are not really relevant.

    However, what is particularly frustrating is that I get an answer of 200, because we need to select 3 rows from 6 and 3 columns from 5, giving a total of ##_3C^6 \, _3C^5 = (20)(10) = 200## distinct choices. At the moment I cannot see anything wrong with either argument! I am hoping my morning caffeine kicks in soon.
  6. Apr 2, 2014 #5


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    That selects the 9 elements in the three rows and three columns, not 3 elements.

    You didn't use the fact that "no two elements are in the same row or column". There are 3! ways of choosing the 3 elements from the 9 (3 choices from the first row, 2 from the second row, 1 from the third), which gives 200 x 6 = 1200 choices.
    Last edited: Apr 2, 2014
  7. Apr 2, 2014 #6
    Ok, yes I think this answers my question! Thanks, all.
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