# Combinations problem

## 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:

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

?