Medium combinatorics
Probability
Level
3
Let's crank up the level!
How many three digit numbers are there such that the first digit is bigger than the last one?
The answer is 450.
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1 solution
Let's start like this: If we want the first number to be 9, then there's 1x10x9 possibilities to make a number whose first digit is bigger than the last (The digit of the middle could be any, and the last one could be any from 0 to 8) If we continue like this, there are <p>1 x 10 x = 90 </p> <p>1 x 10x 8 = 80 </p> <p>1 x10 x 7 = 70 </p> <p>1 x 10 x 6 = 60 </p> <p>1 x 10 x 5 = 50 </p> <p>1 x 10 x 4 = 40 </p> <p>1 x 10x 3 = 30 </p> <p>1 x 10 x 2 = 20 </p> <p>1 x 10x 1 = 10 </p> <p>And if we add the results, the answer is /boxed{450}