1. ## numbers chosen without.replacement

r numbers are chosen at random 0<r<10 without replacement from the numbers 0,1,2,...,9. What is the probability that no two numbers are same ?

2. ## Re: numbers chosen without.replacement

Originally Posted by ice_syncer
r numbers are chosen at random 0<r<10 without replacement from the numbers 0,1,2,...,9. What is the probability that no two numbers are same ?
If you have posted the question correctly, then the probability is one.
Without replacement no two numbers can be the same.

3. ## Re: numbers chosen without.replacement

Hello, ice_syncer!

Please check the wording of the problem.

$r$ numbers are chosen at random $(0 without replacement ??
from the numbers 0, 1, 2, ..., 9.
What is the probability that no two numbers are same ?

The given answer is for with replacement.

Examples:

$P(3) \:=\:\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{1 0} \;=\;\frac{10!}{10^3\,7!}$

$P(4) \;=\;\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{1 0}\cdot\frac{7}{10} \;=\;\frac{10!}{10^4\,6!}$

4. ## Re: numbers chosen without.replacement

That makes sense. It is possible that there is a printing error in my book. Thanks.