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Booze combinations

  1. Jan 23, 2014 #1
    Hi quick question. I was recently posed a question by a co-worker who saw it in a liquor store contest:

    156 different brands of beer for sale
    You can build your own six pack with any of the 156.

    How many different ways can you fill the six pack without having the same brand for more than two bottles?

    My thoughts are that it's just

    156C6 + 156C5 (two of the bottles are one brand = 5 brands chosen) + 156C4 (four of the bottles are from two brands = 4 brands chosen) + 156C3 (two bottles of three brands = 3 brands chosen)

    It's kinda embarrassing that I'm asking this online but I want to make sure I'm not giving him bad advice! Free beer is riding on it. Thanks!
  2. jcsd
  3. Jan 23, 2014 #2

    Simon Bridge

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    If I follow you, you are reasoning that having 1 repeated brand is like having no repeats and only 5 beers to pick?

    That would give you (156x155x154x153x152)/(5x4x3x2) - with only one repeated brand.

    The way to check your reasoning is to look at things another way.

    eg. Lets say I have a 6-hole beer crate.

    If I double up particular brand X, then that leaves me with 4 slots to fill with the rest of the brands.
    So the number of combinations with brand X doubled is the number of combinations of 4 out of the remaining 155 brands.

    There are 156 possible brand X's. So the total number combinations doubling just one brand is:


    which most makes sense?

    And looking another way - you can fill the first 5 holes with 5 individual unique brands, and the remaining hole with one of the previous 5 choices
    That's (156x155x154x153x152)/(5x4x3x2) combinations for the first five holes, times 5 ways of filling the remaining hole.

    ... look familiar?

    Examine each of these approaches carefully and figure which one best applies to your situation.
    It's your beer after all :)
    Last edited: Jan 23, 2014
  4. Jan 24, 2014 #3
    Yes, having 1 repeated brand is like having no repeats and only 5 to pick. Same thing with 2 repeated brands being like no repeats and 4 to pick, 3 repeated brands being like no repeats and 3 to pick.
  5. Jan 24, 2014 #4


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    It doesn't quite work like that when you have a repeated brand. Note that XXABCD is not the same as XAABCD. So, you have more than 156C5. It's actually:

    156(155C4) = 156 choices for the repeat, then 155C4 choices for the others.

    Likewise, with 2 repeats it's


    And, with three repeats it is:


    It's probably best not to think about this again once you've had a few!
  6. Jan 24, 2014 #5

    Simon Bridge

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    So out of the three approaches in post #2, you think the first one makes the most sense?

    Why have you rejected the other two approaches in post #2?

    I'm with PeroK on this one - unless you have left something out of the description, your favorite does not match the problem. Take a closer look at the last approach in post #2 - the result agrees with the second one, but the setup is clearer.

    If you had to pick only 5 bottles, all different, this is exactly having to pick five brands.
    One repeat means you are adding one more bottle to what you have, and, that bottle must be one of the brands already selected - you just said that this makes no difference to the number of possible combinations!

    Simplify the problem so you can count them out easily:
    Pretend there are 4 brands to pick from and four beers to buy, then:

    ABCD are the brands.

    There's only 1 way to get all four unique - just grab one of each.

    One repeat means 3 unique brands + 1 extra.

    ... one possible combination of all-different brands is: ABC
    Repeating one means you could have: AABC or ABBC or ABCC

    ... so for every possible combination of all-different brands, there are three ways to get one repeat.

    Whatever: let us know how you went.
    Last edited: Jan 24, 2014
  7. Jan 24, 2014 #6
    You're both right, thanks for the correction.

    FWIW the contest closed yesterday anyways!
  8. Jan 24, 2014 #7

    Simon Bridge

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    Did you get to enter?
  9. Jan 25, 2014 #8
    It wasn't for me... for a coworker. He just missed the deadline.
  10. Jan 25, 2014 #9

    Simon Bridge

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    Shame - did you get the official answer off the coworker or is it moot?
  11. Feb 1, 2014 #10
    It was in the 21 billion range... what you all said.
  12. Feb 2, 2014 #11

    Simon Bridge

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    I get:

    6 unique brands > 156*155*154*153*152*151
    ans = 1.3076e+13

    1 dupe = 5 unique + 1 out of the 5 > 156*155*154*153*152*(5)
    ans = 4.3299e+11

    2 dupes = 4 unique + 2 out of the 4 > 156*155*154*153*(4*3)
    ans = 6.8367e+09

    3 dupes = 3 unique + 3 out of the 3 > 156*155*154
    ans = 3723720

    total: 1.3516x1013

    i.e. 1.3516x1013 possible combinations of 156 brands in a six-pack, with no brand repeated more than once.

    Unless I've misread the problem.
  13. Feb 2, 2014 #12
    The official answer was the approach PeroK took, which I entered into R as:
    > choose(156,6)+156*choose(155,4)+(choose(156,2)^2)+choose(156,3)
    [1] 21916772956 = 21.9 billion
  14. Feb 2, 2014 #13

    Simon Bridge

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    @PeroK: troubleshoot my reasoning?
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