3-digit numbers greater than 500 must not start with 3 so = 4P3 - 3P2 = 18
The digits 3, 6, 7, and 8 will be used without repetition
to form different 3-digit numbers.
Of all such numbers, how many are greater than 500?
The answer is 18 but don't know how to obtain it.
I don't know if we can help you.
This is definitely a Permutation problem.
. . But you've never heard of Permutations, ?
What grade are you in?
367,368,376,378,386,387 ; there's 3 more similar arrangements...
Btw, no repetition means no number like 777 or 778; all digits must be different...
Go here to "learn":
Math Forum: Ask Dr. Math FAQ: Permutations and Combinations
In high school Soroban and yes have done plenty of permutations--now I know what the "P" is from the post above but not sure why to choose those numbers to form the permutation.
Wimer--the concrete numbers like that help, would the 3 other arrangements consist of numbers starting with 6, 7, and 8? (that is, numbers in the 600s, 700s, and 800s?)
The first number can be either 6, 7, or 8- three choices. The second number can be either of the two remaining digits and 3- three choices again. There are then 2 choices for the third digit. A total of 3*3*2= 18 possible numbers.