# Combinations problem

1. May 4, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?

2. Relevant equations

$$C_n^k=\frac{n!}{k!(n-k)!}$$

3. The attempt at a solution

6 tickets from 10, we can choose on $C_1_0^6$. There are 4 tickets left. [itex]C_7^3[/tex] is the tickets which are not winning. Is the right answer:

$$C_1_0^6$$ - $$C_7^3$$

?

2. May 4, 2008

### mutton

Focus on the "at least" part.

3. May 4, 2008

### Physicsissuef

Sorry, but I cant figure out