Question from AMC10A 2010

Someone posted a question (#18) from AMC10A on the cbox. It's easier and less messy to explain it here rather than on the cbox:

Qn: Bernardo randomly picks 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 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:
Consider two scenarios: (a) Bernardo picks a '9' among the 3 numbers that she picks from her set of 9 numbers (b) Bernardo does not pick any '9'

Bernardo picks ‘9’:
When picking 3 numbers from the set (1,2,3,4,5,6,7,8,9), probability that Bernardo would pick a ‘9’ is 3/9 or 1/3. When this happens, Bernardo’s number (which has to be 9**, since the 3-digit number is to be arranged in descending order) will always be greater than Silvia’s number (obvious, since Silvia's greatest possible 3-digit number can only be 876!).

Bernardo does not pick ‘9’:
Number of different 3-digit numbers: 8C3 = 56
Probability that both Bernardo and Silvia pick the same 3-digit number is 1/56
Probability that both Bernardo and Silvia do not pick the same 3-digit number is 55/56
If both Bernardo and Silvia pick different number, probability that Bernardo’s number is bigger than Silvia’s number is ½ (similarly, probability that Bernardo’s number is smaller than Silvia’s number is also ½)

Therefore the probability that Bernardo’s number is larger than Silvia’s number is:
(Bernardo picks ‘9’) 1/3 + (Bernardo does not pick ‘9’) 2/3 x ½ x (55/56)
= 37/56