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Let's say I have $20$ different pairs of shoes in my wardrobe. I want to know, in how many ways can I choose $8$ shoes, so that I have at least two pairs?

All I know is that I can choose $8$ shoes from $40$ ($20 \cdot 2$) in $\dfrac{40!}{8!(40-8)!}$ ways.

Ken
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Dargenn
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1 Answers1

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These problems are usually easier to solve looking at the reciprocal case. Let's see the probability $p$ of getting $8$ shoes such that there is $0$, or $1$ pairs, then what you are looking for will be $1-p$.

The probability of getting $0$ pairs can be calculated by: for the first shoe, $40$ posibilities, for the second shoe $38$ (we eliminate the pair of the shoe we chose), then $36$, and so on:

$$p_0=\frac{40\cdot 38\cdot 36\cdot 34\cdot 32\cdot30\cdot28\cdot26}{8!P}$$

where $P$ is the total number of posibilities that you have correctly calculated, and $8!$ compensates for the posible permutations of the already selected shoes.

To get only $1$ pair, we have to choose a pair, there are $20$, and of the remaining $38$ shoes, do the same as before to choose the other $6$ shoes, so the probability is:

$$p_1 = \frac{20\cdot 38\cdot 36\cdot 34\cdot 32\cdot 30\cdot 28}{6!P}$$

The probability you're looking for is:

$$1-p_0-p_1\approx 0.13$$

MyUserIsThis
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  • But p0 is much higher than P, and how can probability be more than 1? – Dargenn Jun 11 '15 at 15:18
  • @Dargenn I'm sorry, $P$ should be $P=40\cdot39\cdot38\cdot\ldots\cdot33$, and this solution considers a different case the same set of shoes dissordered, so it's incomplete. You can divide by the number of permutations of the $8$ chosen shoes. – MyUserIsThis Jun 11 '15 at 15:37
  • @Dargenn My result is $p=0.43$, and so the solution would be $0.57$. – MyUserIsThis Jun 11 '15 at 15:42
  • Well, I don't need the probability, but the number of combinations. That's not the problem though, since I can easily get that number now. Thanks! – Dargenn Jun 11 '15 at 16:53