Find the number of five-digit multiples of $5$, where all the digits are different, and the second digit (from the left) is odd.

Question

Find the number of five-digit multiples of $5$, where all the digits are different, and the second digit (from the left) is odd.

Answer 1

Here as the number should be divisible by 5, last digit must be either 0 or 5

So consider two cases...

• case 1:-last digit is 0
  In this case, you will have now 9 digits left (as all the digits are different). Also you can fill second place using one of 5 odd digits out of remaining 9 digits. Now start filling from 1st place, which can be filled in 8, 3rd in 7 and 4th in 6 ways.

So in total you will have (8)(7)(6)(5) = 1680 numbers of case 1.

• case 2:- last digit is 5
  In this case you have now 4 odd digits and 5 even digits including 0. So now you can fill second place with remaining 4 odd digits. Now you are left with 3 odd digits and 5 even digits. So starting from 1st place, you can fill 1st place in 7 ways(you can't fill 1st place with 0), 3rd place with 7 ways(now you can include 0) and 4th place with 6 ways...

So in total in case 2 you have (7)(4)(7)(6) = 1176 numbers.

So, total of case 1 and case 2 will give you 2856 numbers.