logann2 wrote:

Carcass wrote:

This is how Edward’s Lotteries work. First, 9 different numbers are selected. Tickets with exactly 6 of the 9 numbers randomly selected are printed such that no two tickets have the same set of numbers. Finally, the winning ticket is the one containing the 6 numbers drawn from the 9 randomly. There is exactly one winning ticket in the lottery system. How many tickets can the lottery system print?

(A) \(9P_6\)

(B) \(9P_3\)

(C) \(9C_9\)

(D) \(9C_6\)

(E) \(6^9\)

9 numbers, choose 6 of them. 9C6. D.

Selected exactly 6 of the 9 available numbers means they're randomly picking 6 of the 9 numbers. The part that lets you know that it's indeed a choose function and not permutation is that the problem says that "no two tickets contain the same numbers" and (paraphrasing) "the winning ticket contains the 6 winning numbers"

So order doesn't matter. Does that help?