- #1

LCSphysicist

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- 162

- Homework Statement
- > 20 shoes, from 10 pairs of shoes, are lined randomly. What is the

> probability that there is a set of 10 consecutive shoes with 5

> left shoes and 5 right shoes?

- Relevant Equations
- .

> 20 shoes, from 10 pairs of shoes, are lined randomly. What is the

> probability that there is a set of 10 consecutive shoes with 5 left shoes

> and 5 right shoes?

I thought that would be a good idea to imagine 10 shoes as one unique object, as follow:

Instead enumerate 20 objects, let's separate 10 shoes and think them as another object (call it X), so that there is 11 in total.

There is 11! ways to permute each objects. The problem is to know how to count X.

So i am stuck in how to count |X| right. I thought that we could first choose ##C_{10,5} = 10!/(5!5!)## right shoes, ##C_{10,5}## left shoes. Now, the number of ways that satisfy our conditions is ##C_{10,5}*C_{10,5}*\frac{{10!}*11!}{1}##, so the prob is ##C_{10,5}*C_{10,5}*\frac{{10!}*11!}{20!}##. But this is greater than 1 (...), i have no idea what to do now.

I would appreciate to receive an answer that specifies where is my error :.

> probability that there is a set of 10 consecutive shoes with 5 left shoes

> and 5 right shoes?

I thought that would be a good idea to imagine 10 shoes as one unique object, as follow:

Instead enumerate 20 objects, let's separate 10 shoes and think them as another object (call it X), so that there is 11 in total.

There is 11! ways to permute each objects. The problem is to know how to count X.

So i am stuck in how to count |X| right. I thought that we could first choose ##C_{10,5} = 10!/(5!5!)## right shoes, ##C_{10,5}## left shoes. Now, the number of ways that satisfy our conditions is ##C_{10,5}*C_{10,5}*\frac{{10!}*11!}{1}##, so the prob is ##C_{10,5}*C_{10,5}*\frac{{10!}*11!}{20!}##. But this is greater than 1 (...), i have no idea what to do now.

I would appreciate to receive an answer that specifies where is my error :.