Find the number of five-digit multiples of 5, where all the digits are different, and the second digit (from the left) is 1.
For the first digit, there are 9 choices (0 - 9)
For the second digit, there is 1 choice (it has to be 1)
For the third digit, there are 10 choices ( 0 - 9)
For the fourth digit, there are 10 choices (0 - 9)
For the final digit, there are 2 choices (0 - 2)
So, there are \(9 \times 10^2 \times 2 = \color{brown}\boxed{1,800}\) choices.
Two cases
Case 1 - Last digit = 0
We have eight choices for the first digit (2,3,4,5,6,7,8,9)
One for the second (1)
Seven for the third
Six for the fourth
8 * 1 * 7 * 6 = 336
Case 2 - Last digit = 5
We have seven choices for the first digit ( 2,3,4,6,7,8,9 )
One for the second
Seven for the third
Six for the fourth
7 * 1 * 7 * 6 = 294
336 + 294 =
630 such numbers