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Digit Problem on Probability

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Digit Problem on Probability

by Preeti.a8 » Tue Aug 07, 2012 10:31 pm
Hi ,
I have a probability problem and would want to know the solution .

Problem:-
I want to create all 5 digit numbers with the below condition.
The 1st digit should be a even number. The second digit should be a odd number.
The 3rd digit should be a non even prime number. Fourth and fifth should be random digits which do not occur previously.

As per the solution the answer given is 4(even numbers)* 5( odd numbers) * 3 * 7 * (non used digits) * 6 (non used digits) which is 2520.
But here the cases have not been considered when 2nd and 3rd digit can be the same ie the below cases (3,3) , (5,5) and (7,7). In these cases the 4th and 5th numbers should be 8(non used digits)*7(non used digits) . Since there are 3 such combinations we multiply it by 3 ie 56*3=168.

Could anyone let me know if (2520+168) is correct or the answer 2520 is the right one
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by Pharo » Wed Aug 08, 2012 1:03 am
2688 is the correct solution :)
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by Preeti.a8 » Thu Aug 09, 2012 12:16 am
Thanks . Could you let me know some refrence where you found the solution??
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by Pharo » Thu Aug 09, 2012 2:46 am
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

---

Hope that helps :)
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by Preeti.a8 » Fri Aug 10, 2012 12:51 am
Thanks :)...well explained solution ..
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by Pharo » Sat Aug 11, 2012 4:25 am
No worries :)
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