1. ## Forming 3-digit numbers

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

2. ## Re: Forming 3-digit numbers

3-digit numbers greater than 500 must not start with 3 so = 4P3 - 3P2 = 18

3. ## Re: Forming 3-digit numbers

I'm still confused--I understand why they cant start with 3 (needs to be more than 500) but I don't understand the repetition part. Also, what does the last part mean with the "P"

4. ## Re: Forming 3-digit numbers

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.

This is definitely a Permutation problem.
. . But you've never heard of Permutations, $_nP_r$ ?

5. ## Re: Forming 3-digit numbers

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

6. ## Re: Forming 3-digit numbers

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?)

7. ## Re: Forming 3-digit numbers

Originally Posted by dannyc
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?)
Of course!

8. ## Re: Forming 3-digit numbers

So then there would be 6 numbers to those 3 sets as well? (637,638,673,678,683,687) Would those be the correct ones for the 600s?

9. ## Re: Forming 3-digit numbers

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.

10. ## Re: Forming 3-digit numbers

Thanks Ivy!! That makes much more sense! (I should've thought of that)

11. ## Re: Forming 3-digit numbers

Originally Posted by dannyc
So then there would be 6 numbers to those 3 sets as well? (637,638,673,678,683,687) Would those be the correct ones for the 600s?
Danny boy, why ain't you SURE those are correct?

12. ## Re: Forming 3-digit numbers

LOL I guess I just needed that confidence boost. Thanks!