Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A k-element selection, repetitions allowed, with a twist

  1. Feb 7, 2013 #1
    Hey I have kind of a general question.

    In my text book, there are lots of questions that go something like this:

    "Say we have 3 piles of balls, a blue pile, a red pile, and a yellow pile. Each pile has at least 8 balls. How many ways can we pick 8 balls from these three piles?"

    (so for example in this case you could pick YYYYYYYY or BYYRRRRR, etc.)

    In this case we use C(k+t-1,t-1) formula

    But I'm stumped on ideas like this:

    What if I take the above scenario, but this time the "blue pile" has only 2 balls, while the rest have 8. And I'd still like to pick 8 balls. I can't use the formula like normal because there are less options: for example BBBYYRRR isn't an answer as there are only 2 blue balls in the pile. Now let me go even further and say, there are only 2 blue balls, and only 3 red balls and only 4 yellow balls and I want to make a selection of 8 balls. I am completely stumped as to how to do this as I can't apply the normal formulas.

    So I guess, is there a formula for when one of your sets is "limited"? Every single problem in the book the example is to make a k-element selection but the "sets" you are choosing from always have at least k elements in their set so you don't run in to this problem.

    This is NOT a HW problem, this is just something I keep trying to figure out on my own. I am really, really poor at combination/permutation stuff so there is most likely a very easy answer to this I'm just not seeing.
  2. jcsd
  3. Feb 7, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    In that case you have to count them out rather than use the formula.

    If you don't care which colors are included, then it is just the number of ways of picking eight balls from 8+8+2 balls. For anything else you need to be specific - eg: perhaps you must have at least one ball of each color?

    In that case, you are selecting from 7+7+1 balls in five trials (the first three are certain).
    Does the order matter? Stuff like that. But you should have got an idea how to think about these now.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook