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Even number set { 0, 2, 4, 6, 8 }   Odd number set  { 1, 3, 5, 7, 9 } 

The answer can be achieved by considering 2 cases

1. 3 digits are even and 1 digit is odd

2. 3 digits are odd and 1 digit is even

1st case: 

  3 digits are even and 1 digit is odd 

      here 2 cases are possible     

  • 2 non-zero even digits and 1 digit = 0 and 1 odd digit​​​​​​

      OR

  • 3 non-zero even digit and 1 odd digits ​​​​​​​

---------------------------------------------------------------

  • 2 non-zero even digits and 1 digit = 0 and 1 odd digit​​​​​​

           Choosing 2 non-zero even digit = 4C2 = 6

           Choosing 1 odd digit = 5C1 = 5

           Arranging all 4 digits = 3 × 3 × 2 × 1 = 18

           The no. of ways will be = 18 * 6 * 5 = 540    [the numbers can't be started with 0] 

  • 3 non-zero even digit​​​​​​​ and ​​​​​​​1 odd digits ​​​​​​​

           Choosing 3 non-zero even digit = 4C3 = 4

           Choosing 1 odd digit = 5C1 = 5

           Arranging all 4 digits = 4! = 24

           The no. of ways will be = 24 * 5 * 4 = 480

The total no. of ways of 1st case = 540 + 480 = 1020

Now the 2nd case

3 digits are odd and 1 digit is even

  • 1 digit = 0​​​​​​​ and 3 odd digits 

       OR

  • 1 non-zero even digit​​​​​​​ and 3 odd digits 

-----------------------------------------------------

  • 1 digit = 0​​​​​​​ and 3 odd digits 

           Choosing 3 odd digits = 5C3 = 10
           Arranging all 4 digits = 3 × 3 × 2 × 1 = 18

           The no. of ways will be = 18 * 10 = 180 

  • 1 non-zero even digit​​​​​​​ and 3 odd digits 

           Choosing 1 non-zero even digit = 4C1 = 4

           Choosing 3 odd digit = 5C3 = 10
           Arranging all 4 digits = 4! = 24

           The no. of ways will be = 24 * 4 * 10 = 960

The total no. of ways of 2nd case = 180 + 960 = 1140

The total no. of ways of 1st case + The total no. of ways of 2nd case = 1020 + 1140 = 2160 ways

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