Talk:2010 AMC 12A Problems/Problem 16

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Hi!! I need help with AMC 10A 2010 Problem #18. Here's the problem and solution, as well as what I don't get at the end:

Bernardo randomly picks 3 distinct numbers from the set [1,2,3,4,5,6,7,8,9] and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1,2,3,4,5,6,7,8} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

Solution We can solve this by breaking the problem down into 2 cases and adding up the probabilities.

Case 1: Bernardo picks 9. and then it says the solution

Case 2: Bernardo does not pick 9. Since the chance of Bernardo picking 9 is 1/3, the probability of not picking 9 is 2/3.

If Bernardo does not pick 9, then he can pick any number from 1 to 8. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.

Ignoring the 9 for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.

We get this probability to be 3!/8*7*6

THIS IS WHAT I DON'T GET!! Why wouldn't it be 8 choose 3???

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