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Lets assign ABC for three places
A can have 8 values
Lets say b is 0, C can have 10 values
Total = 8*10 = 80
Similarly when B is 9 , c can have only 8 values since 9 is not possible
Total = 8*9 = 72

Total ways = 80+72 = 152

Answer A

Posted from my mobile device


Hello, good explanation, thank you! I just noticed one small typo: When B is 9, C can have only 9 values (not 8).[/quote

Thanks mate. Overlooked.
It was a typo. Glad you noticed it !
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My 2 cents on this:

The conditions are a great way to learn about arrangements. Based on the question there are a few conditions. Firstly, 0 to 9 equals 10 digitals. Remember that. Easy to take this for granted.

Firstly, the first digital possibilities are fixed as 8 given that both 0 or 9 cannot be the first digit (10-2)

Then on to the cases:

Case 1: when 2nd digit is 0

8 x 1 x 10 (1 being value of only 0, and 10 being all of the ones we can apply, including 9) = 80

Case 2: When 2nd digit is not 0, then 3nd and 3rd cannot both be 9

8 x 1 x 9 (1 being value of9, and since we have taken 9 as 2nd, then third cannot have a 9, so 10-1 = 9 ways) = 72

80 +72 = 152

A
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