A Question of Probabilities and Combinations

Discussion in 'Science and Technology' started by ThunderAeroI, Dec 16, 2016.

  1. ThunderAeroI

    ThunderAeroI Rear Admiral Rear Admiral

    Joined:
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    Location:
    Perpetually being chased by airplanes
    Lets say you have seven lists, and each list has a different number of elements:

    L1: 3
    L2: 2
    L3: 8
    L4: 10
    L5: 4
    L6: 11
    L7: 2

    How would I determine the maximum number of different combinations that are possible if you must select one and only one element from each list?

    I promise this isn't math homework, its something for a story I am working on.
     
  2. DonIago

    DonIago Vice Admiral Admiral

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    Location:
    Burlington, VT, USA
    Isn't that just 3 x 2 x 8....
     
  3. Asbo Zaprudder

    Asbo Zaprudder Admiral Admiral

    Joined:
    Jan 14, 2004
    Location:
    Rishi's Sad Madhouse
    The number of combinations without regard to order for which order you pick from each list = L1 x L2 x L3 x L4 x L5 x L6 x L7 = 42,240 in this case.

    Permutations are a little more tricky as you can pick from the lists in random order. The total number of ways of ordering the lists is N! (N factorial), where N is the number of lists, so in this case 7! = 5,040. So the total number of permutations is 5,040 x 42,240 = 212,889,600.
     
    Last edited: Dec 16, 2016