My solutions to the 2019 Mathematics Extension 2 Paper (1 Viewer)

Nash__

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after looking at these solutions, what do you think 93 will align to?
 

Jimmy9945

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Well how I did it was 10C2(2^4-2).
The 10C2 to get the two numbers, and the 2^4 to choose either of the 2 numbers in each of the 4 spots. The -2 us to exclude the combinations where they are all the same number, eg 2222
 

sharky564

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Isn’t it counting 2211 and then 1122?
As u r choosing 2 spots, and then fit in one spot with a digit, fit in another with another digit
No, what you're saying is you're either choosing any two of the 4 possible spots for one digit (e.g. you're choosing the last two digits to be 1 if your initial 2 digits are 2, 1) and iterating over all possible ordered pairs of numbers, meaning you also count the time you choose the first two digits to be 2, which is identical to the previous case. As a result, you double count these numbers, resulting in too many numbers. I have verified with a Python program and it has also given 630 as the answer:

Python:
count = 0
numbers = [str(a) + str(b) + str(c) + str(d) for a in range(0, 10) for b in range(0, 10) for c in range(0, 10) for d in range(0, 10)]
for x in numbers:
    if len(set(list(x))) == 2:
        count += 1

print(count)
 

psmao

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Well how I did it was 10C2(2^4-2).
The 10C2 to get the two numbers, and the 2^4 to choose either of the 2 numbers in each of the 4 spots. The -2 us to exclude the combinations where they are all the same number, eg 2222
Wouldn’t u need 10P2 coz these two numbers can swap their positions
 

psmao

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What would 96 align to
I think I fked up my state rank holly sht
 

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