How many different 3 digit numbers can be made from the digits 1 to 7 if repetition is not allowed and the number must be an even number less than 200

Question

i am unsure about the steps taken to get the answer

Answer 1

Okay, the number must be 3 digits and less than 200.  That means the first digit is 1. (one possibility)

 

The number must be even.  That means the third digit is 2,4, or 6. (Three possibilities)

 

There are no repeats, so the second digit must be among the set (2,3,4,5,6, 7), but not the one we have already used for the third digit.  Five possibilities for each third digit.

 

Total combinations, 1 times 3 times 5, for a total of 15. 

 

Answer 2

Here's how I was shown how to do problems like this:

 

Answer this question for each digit...how many different numbers can you have?

 

For the first digit, you can have 7 different possibilities (any of the numbers 1-7)

 

For the 2nd digit, since they cannot repeat, you only have 6 different possibilities (one of the numbers is being used for the first digit).

 

For the 3rd digit, you only have 5 possibilities (two of the numbers are already in use).

 

Then you just multiply these numbers together to get the total number of possible 3 digit numbers.

I get 210, so I'm not sure how it's less than 200.

 

I even tried it manually and got 30 different possibilities for each different number in the first digit spot.  Since it can be 7 different possibilities, 30*7 = 210.