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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
avatar+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
avatar+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

 

 

cool cool cool

 Jul 7, 2022

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