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A jar contains 17 red balls and 22 blue balls. How many ways are there to choose, without replacement, 8 balls from this jar.

I have two answers, but they both seem right to me. Could some explain which one is the wrong answer and why?

  1. Since we choose 8 balls, it means that we are choose a combination of balls from each color. I can sum up the different ways to calculate the different amounts through distributed counting. $\sum_{k=0}^8 \left({17 \choose k} + {22 \choose 8-k}\right)$

  2. Another way to look at this is the pool the balls together. I know there are 39 balls in total, so I choose the combinations to pick 8 balls from 39. ${39 \choose 8}$

Which logic makes more sense?

MT_
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GivenPie
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  • Forgive me if this is implicit, but what exactly do you mean by "choose?" If we only care about the end product (how many red and blue balls we have), then the answer is $9$. That's what comes to my mind when it says "choose," but clearly you are thinking of something else. – MT_ Apr 17 '14 at 02:36
  • Choose, as in the different combinations on red to blue balls if we took a subset of 8 different balls from this pool of balls. – GivenPie Apr 17 '14 at 02:38
  • So are the red balls distinct to each other? If the red (and blue) balls are indistinct, then the answer would be $9$, which is how I would interpret the problem. – MT_ Apr 17 '14 at 03:57

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Both logic are correct, they will give the same answer. To convince yourself check on a smaller example.

  • Not quite, in first formula it should be $\sum \binom{17}{k}\binom{22}{8-k}$. – André Nicolas Apr 17 '14 at 02:27
  • I guess that's what GivenPie wrote (intended to write) –  Apr 17 '14 at 02:29
  • @AndréNicolas I thought by adding the different distributions of the selection, we can find all the amounts between red and blue. Thanks for clarifying Andre. – GivenPie Apr 17 '14 at 02:33
  • Yes, "outside" you add. But for example the probability of $3$ red and $5$ blue is $\binom{17}{3}\binom{22}{5}$, since for every way of choosing $3$ red there are $\binom{22}{5}$ ways to choose $5$ blue. – André Nicolas Apr 17 '14 at 02:39