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A Problem with Permutations

  1. Dec 19, 2015 #1
    Hey guys , Could anyone here tell me the easiest way to solve for n , nP7 = 604800 , the traditional way (I'm currently using) is to divide 604800 by 10 and then 9 and so on until I get 1 as a result of that division , The problem is this way isn't helpful with all permutations I have in my study , some times it gets tricky .
     
  2. jcsd
  3. Dec 19, 2015 #2

    Mentallic

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    Notice that 604,800 is divisible by 100, which means that 10 must be one of the numbers in the factors that multiply to give [itex]^nP_7[/itex]. This is because we only have one prime factor of 5 which can couple with a prime factor of 2 in various ways, such as from 4 or 6 or 8, and that would give a factor of 10, so the other factor of 10 must be from 10 itself, or even 15 (the factor of 5 coupled with another 2 elsewhere). However, 15*14*...*9 is composed of 7 factors that are all mostly larger than 10, so the result is going to be larger than 107 = 10,000,000 which is much too large. So we only have a few to test, starting at 10*9*...*4 and working our way upwards from there. We actually find that n=10 gives the answer, but if for argument's sake we had to solve, say,

    [itex]^nP_7=3,991,680[/itex]

    Then we know that n=15 is again too large, but now it's also not divisible by 100 any more, which means we're missing a prime factor of 5, and how can that happen? Only with
    12*11*...*6
    13*12*...*7
    14*13*...*8
    n being one larger or smaller than these possibilities and we'd end up with the factor of 5 in 5 or 15 and the result would be divisible by 100. So given just a few values to test, the result should be quick to find.
     
  4. Dec 19, 2015 #3
    Thanks for your help :D
     
  5. Dec 19, 2015 #4

    Mentallic

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    You're welcome!
     
  6. Dec 19, 2015 #5

    WWGD

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    How about finding the factorization of 604800 as a product of primes? Then divide by 2!, then by 3! , etc.
     
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