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
Hello, dannyc!
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, $\displaystyle _nP_r$ ?
What grade are you in?
HINT:
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.