If one wants to find the probability of winning the lottery and the lottery asks you for 8 numbers from 58 numbers and an extra one to choose from another 22 numbers. What is the probability of winning? Do I use permutations or combinations?
Note: the order of the numbers does not matter. The numbers can also repeat. I've tried using combinations. This is my way of doing it: $$C(8,58)=\frac{58!}{8!50!}=\frac{1}{8!}\frac{58!}{50!}= 1916797311\cdot22=42169540842$$ Is it correct?
The mnemotechnic I've been taught is "if position matters, than it's permutaion". Since the position (order) of the numbers doesn't matter it's definitely combinations- in how many ways can you combine $n$ numbers into a set of $k$ elements, with repetition - this number is given by ${n+k+1}\choose{k}$. Now, to take into account that you also need the guess the additional number, you additionally multiply by $22$.
In my answer I assumed the numbers can repeat - so for example eight $1$'s with an additional $1$ is a valid lottery ticket. If withing the first $8$ numbers you choose there is no repetition (as is the case with most lotteries, because just imagine how a lottery ticket with repetition would look like, you'd have to choose how many of each number you take, instead of just selecting it), then indeed your formula is correct.
