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## 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

[tex]C_n^k=\frac{n!}{k!(n-k)!}[/tex]

## The Attempt at a Solution

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

[tex]C_1_0^6[/tex] - [tex]C_7^3[/tex]

?