Problem Solving

Codes must be formed having five characters comprising 3 letters and 2 digits.If all the digits are available and only 20 alphabets must be used,then how many different codes can be formed if every code can have as many alphabets and digits possible.

Thanks in advance for the answers with explanations.

The answer is [spoiler] :(20 x 20 x 20 x 10 x 10 x 10) x 5C3 (tricky part) ways.[/spoiler]

naveendevaraj wrote:Codes must be formed having five characters comprising 3 letters and 2 digits.If all the digits are available and only 20 alphabets must be used,then how many different codes can be formed if every code can have as many alphabets and digits possible.

Nice problem... I will give my shot!

Let us consider each "A" a place to choose one between 20 alphabets and "D" a place to choose one between 10 digits.

(Phase 1) How many configurations are possible to put in sequence 3 A´s and 2 D´s ?

This is a permutation with repetitions, so I suppose you will understand that the answer for this is 5! over 3!2! , that is, 10.

(Phase 2) Let us choose and fix any of the 10 configurations obtained in Phase 1, say, "A A A D D". For this configuration, AS FOR ANY OTHER of the 10 available(*), you have 20^3 * 10^2 possible choices.

(Phase 3) Applying the Multiplicative Principle (validaded by (*) and realizing there are no "double counting"´s), I believe the answer to your question is 10 (20^3 *10^2) = 20^3 * 10^3 = 200^3 = 8*10^6.

The answer is correct? If so, do you have any doubts about my solution?

Best Regards,
Fábio.

thanks Fabio for your explanation

naveendevaraj wrote:thanks Fabio for your explanation

My pleasure!

P.S.: the answer is exactly the same as mine, just for the record, but I agree that the "C(5,3) argument" is easier to grasp than the "permutation with repetition" I mentioned.

For the ones who would like to compare both approaches, I leave here the following problem:

How many anagrams are there for the word ARARA... (This is a beautiful bird quite common in Brazil, by the way)

(a) Doing it through COMBINATIONS + PERMUTATIONS ?

(b) Doing it through "permutations with repetitions" (only) ?

Hope you like it!

naveendevaraj wrote:Codes must be formed having five characters comprising 3 letters and 2 digits.If all the digits are available and only 20 alphabets must be used,then how many different codes can be formed if every code can have as many alphabets and digits possible.

Thanks in advance for the answers with explanations.

The answer is [spoiler] :(20 x 20 x 20 x 10 x 10 x 10) x 5C3 (tricky part) ways.[/spoiler]

hey, may i know what is the source?

This is how i figured out the answer :

there are 3 letters and 2 digits , how many ways can 3 lettrs and 2 digits be arranged?

so I mentally imagined doing this : LDLDL or LLLDDD or DLDLL etc..

now for each arrangement I calculate the number of ways to make the code

so LDLDL = 20*10*20*10*20
LLLDDD = 20*20*20*10*10 etc..

So for each arrangement the number of ways to arrange the letters and digit is 20^3*10^2

so if i find out how many ways 3Letters and 2 digits can be arranged I can multiply it to 20^3*10^2

Number of ways to arrange is 5! with repetitions of 2 digits and 3 letters , 5! divided by 2! and 3!

So answer is 20^3*10^2 * 5!/3!*2!

experts I guess the above given solution is incorrect
here is the explaination

20x19x18x10x9x 5! + 20x1x19x10x9x 5!/2! + 20x1x19x10x1x 5!/2!x2! + 20x1x1x10x9x 5!/3! + 20x1x110x1x 5!/3!x2!

experts any clue ??

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.

awesome explaination,
considerind the two flaws in my explaination, I got
20x19x18x10x9x 5!/3!x2! + 20x1x19x10x9x 5!/2! +
20x1x19x10x1x 5!/2!x2! + 20x1x1x10x9x 5!/3! +
20x1x1x10x1x 5!/3!x2! + 20x19x18x10x1/2!(missed case)

6156000 + 2052000 + 114000 + 36000 + 2000 + (4104000 the missed case) = 12464000 which is not same as ans in original post " 8x10^6 "

The other parcels are also flawed. (I insist: study the much simpler solutions presented, and feel free to ask about them if needed.)