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# help math

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36
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Find the number of five-digit multiples of 5, where all the digits are different, and the second digit (from the left) is 1.

Jul 6, 2022

#1
+2293
+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.

Jul 6, 2022
#2
+124526
+1

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

Jul 7, 2022