Digit Problem on Probability

2688 is the correct solution

Thanks . Could you let me know some refrence where you found the solution??

I solved it myself

Here is my solution:

Digit 1 can be {2,4,6,8}
Digit 2 can be {1,3,5,6,7}
Digit 3 can be {3,5,7}
Digit 4&5 are dependent on previous digits.

Case 1: Digit 2 and 3 are same (and belong to the subset {3,5,7})

Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {3} --> 1 option
Digit 3 can be {3} --> 1 option
Digit 4 can have 10-2 = 8 options
Digit 5 can have 10-3 = 7 options
---- total = 4*1*1*8*7 = 4*56 options

You can repeat case 1 three times for each 3,5 and 7 (i.e. digit 2 can be 3, 5 or 7 etc..).
Hence total options from case 1 = 3*4*56 = 12*56

Case 2: Digit 2 and 3 are different (and belong to the subset {3,5,7})

Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {3} --> 1 option
Digit 3 can be {5,7} --> 2 option
Digit 4 can have 10-3 = 7 options
Digit 5 can have 10-4 = 6 options
---- total = 4*1*2*7*6 = 6*56 options

You can repeat case 2 three times for each 3,5 and 7. (i.e. digit 2 can be 3, 5 or 7 etc..)
Hence total options from case 1 = 3*6*56 = 18*56

Case 3: Digit 2 belongs to the subset {1,9} and Digit 3 belongs to the subset {3,5,7}

Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {1,9} --> 2 option
Digit 3 can be {3,5,7} --> 3 option
Digit 4 can have 10-3 = 7 options
Digit 5 can have 10-4 = 6 options
---- total = 4*2*3*7*6 = 18*56 options

There is no repetition for this case. Total = 18*56

Grand total of possibilities are the sum of all the above totals:

GT = 12*56 + 18*56 + 18*56 = 48*56 = 2688

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Hope that helps

Thanks ...well explained solution ..

No worries