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 ?
Answer is 10!/((10^r)*(10-r)!)
please explain reasoning
Hello, ice_syncer!
Please check the wording of the problem.
numbers are chosen at random without replacement ??
from the numbers 0, 1, 2, ..., 9.
What is the probability that no two numbers are same ?
Answer is: 10!/[(10^r)*(10-r)!]
The given answer is for with replacement.
Examples: