Hi vipulgoyal,
It´s a pleasure to come to this forum after a very long time.
First of all, if all answers/solutions were right, I guess it is easy to realize that yours is the "harder" one by far, therefore I suggest you try to find easier ways, not only because they take much less time (this is important), but also because they will have less "chance" to trick you, so to speak.
That understood, I will NOT show you how to fix your solution from the beginning to the end, simply because it takes too much time. I will simply show you two things:
1) your "cases" are not counted properly, and I will show where you committed a mistake, in detail, in your FIRST one.
2) you forgot one scenario, I mean, you should work SIX cases, not five as you did;
Before doing that, let me suggest you study the solutions shown above, to be sure you understand why they are right. If you think something is not done properly, please feel free to ask till the moment you understand there is really nothing wrong (well, I guess so)!
Ok, so let´s see the topics mentioned above:
TOPIC 1:
Let´s look at the first parcel presented in the sum you presented:
> 20x19x18 admits all chosen letters are different
> 10x9 admits the 2 chosen digits are different
> 5! is the number of permutations of the 5 distinct objects
Wrong partial answer: (20x19x18) (10x9) 5! = 73,872,000 --> This is HUGE!!
[Explanation of the mistake]:
One of the 20x19x18 letter options is (A,B,C) and one of the 10x9 digit options is (0,1), creating the (A,B,C,0,1) choice.
Another of the 20x19x18 letter options is (B,A,C), and another of the 10x9 options is (1,0), creating the (B,A,C,1,0) choice.
(Yes, (A,B,C,0,1) and (B,A,C,1,0) are really different choices, so far so good...)
The problem is that permutating the (A,B,C,0,1) choice you will get the (B,A,C,1,0) choice and vice-versa, therefore when you multiply (20x19x18)(10x9) by 5! you are creating double-counting´s, because the "original" (A,B,C,0,1) will create the (B,A,C,1,0) "copy" while its "original version" was already counted BEFORE the multiplication by 5!
[Explanation for the correction]:
(20x19x18) runs over all "letters scenarios" and (10x9) runs over all "digit scenarios", therefore you need to "filter" to (20x19x18)/3! = C(20,3) and (10x9)/2! = C(10,2) before multiplying by 5! = P(5) , because doing that you will be sure NOT to take into account the relative order between letters and between the digits, right before you WILL take into account the relative order of all the 5 distinct objects!
Right partial answer: (20x19x18)/3! * (10x9)/2! * 5! = C(20,3) * C(10,2) * P(5) = 6,156,000
--> Please note this number is really smaller, much smaller, than the one you got!!
TOPIC 2:
The cases ("parcels presented at the sum") number (1), (2) and (4) deal with all possible scenarios where the 2 digits are different (with 3 different letters, two repeated and one different and finally just one letter repeated three times), but cases (3) and (5) are just two of the three cases where the 2 digits are the same, therefore we realize that there is one parcel missing, that is, the parcel related to 2 digits that are equal and three different letters. You simply forgot this scenario!!
Well, I hope you enjoy my explanations!
Regards and very good luck in your studies,
Fabio.
