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Thread: Discreete/Combinatorial math problem

  1. #1 Discreete/Combinatorial math problem 
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    Suppose that a weapons inspector must inspect each of five different sites twice,
    visiting one site per day. The inspector is free to select the order in which to visit these sites, but cannot visit site X, the most suspicious site, on two consecutive days. In how many different orders can the inspector visit these sites?

    I am assuming that first I would consider the "inspect each of five different sites twice" just as 10 different sites to visit. The number of possibilities for that would be 10!. Then I would substract from that the number of cases where site X is visited 2 days in a row. I am stuck on how to calculate that (the cases where x site is visited on two consecutive days). Any ideas?


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  3. #2  
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    Quote Originally Posted by Lokantis
    The number of possibilities for that would be 10!.
    Not quite. The number of possibilities is .

    Label the five sites V, W, X, Y, Z. Then the total number of visits is precisely the number of anagrams (permutations of the letters) of the word VVWWXXYYZZ.

    Now, with this analogy, can you calculate how many anagrams are there in which XX occurs?


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  4. #3  
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    There is 9 different permutations where 2 X's are consecutive, but each of those possibilities there is 8! permutations for the rest of the sites. So it would be 9 * 8!. Is that right?
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  5. #4  
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    Nope.

    First of all, look at the formula for the total number of visit orders without restrictions:

    Do you know how that is derived? Once you know how that is worked out, you can apply the same reasoning to the next part.
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