Calculating Possibilities for Winning Tickets | Combinations Problem

[Physicsissuef]

Homework Statement

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?

Homework Equations

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

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:$$C_1_0^6$$ - $$C_7^3$$<br /> <br /> ?[/itex]

 

[mutton]

Focus on the "at least" part.

 

[Physicsissuef]

Sorry, but I can't figure out