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SAT I Probability question

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data
    How many different arrangements can be made with these 5 cards?
    There are 5 different cards laid out next to each other. How many different arrangements can be made so that one of the inner cards is never on either end?

    2. Relevant equations
    None, probably

    3. The attempt at a solution
    I know the answer (72), but I have no idea how to get it. It's basic probability, so there should be a simple way of getting it (i.e., no formulas)...

    I tried 252 minus 5 because I didn't know what else to do.

    Can someone help guide me?
  2. jcsd
  3. Jan 17, 2009 #2


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    Science Advisor
    Homework Helper

    Let's do the same problem with the sequence of letters ABCDE. You want all to count all of the arrangements where one of the letters, say B, doesn't occur at either end. Count the number of possibilities for each letter in turn. That means there are 3 different positions B can occupy (since it can't be at either end). That leaves 4 positions to put the A (since the B is in one of the places). Which in turn leaves 3 positions for the C (since A and B are occupying 2 positions). 2 positions for the D, and 1 for the E. Multiply them all together. 3*4*3*2*1=72.
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