finite math

Question

Using digits from this set {1, 2, 3, 4, 5, 6, 7, 8, 9} , how many different 4 digit numbers can be  
      formed if digits cannot be repeated and the 4 digit number must be odd? 

Answer 1

Normally, we'd go with the different possibilities of the 1st number, followed by the 2nd, 3rd, and 4th and multiply to get the total possibilities.

 

BUT, since the last digit has a restriction, let's find the possibilities in reverse.

 

The total possible numbers that can go in the 4th spot is 5  (since there are only 5 digits that would make the number odd)

 

The total possible numbers that can go in the 3rd spot is 8 (not 9 since one of the numbers was used in the 4th spot and numbers cannot repeat)

 

The total possible numbers that can go in the 2nd spot is 7

 

The total possible numbers that can go in the 1st spot is 6

 

SO, the total possibilities is 5 x 8 x 7 x 6 = 1680