All of the stocks on the over the counter market are designated by eit

All of the stocks on the over the counter market are designated by either a 4 letter or a 5 letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be desgnated with these codes?
A. 2 (26)^5
B. 26(26)^4
C. 27(26)^4
D. 26(26)^5
E. 27(26)^5

In 4-digit code {XXXX} each digit can take 26 values (as there are 26 letters), so total # of 4-digits code possible is 26^4;

The same for 5-digit code {XXXXX} again each digit can take 26 values (26 letters), so total # of 5-digits code possible is 26^5;

Total: \(26^4+26^5=26^4(1+26)=27*26^4\).

Answer: C.
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Important point to note here is that letters are not distinct , i.e we can have a code as aaaa or aaaaa for 4 or 5 letter words respectively.


This question is similar to the question
if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
In which we have selected 4 letters from 10 and 5 letters from 10 but in this case the letters have to be distinct.

so using \(P^{10}_{4}\) and \(P^{10}_{5}\)
we get \(\frac{10!}{6!}\) and \(\frac{10!}{5!}\)

But cannot we use the same logic here to select 4 letters from 26 or 5 letters from 26, why?... because the letters are not distinct ( letters can be repeated ) and we cannot use the general permutation formula when there is repetition .

so we cannot use \(P^{26}_{4}\) \(+\) \(P^{26}_{5}\)


if this question were each four letter code and 5 letter code are made of distinct elements then the answer, I think could be
\(P^{26}_{4}\) \(+\) \(P^{26}_{5}\). 4 distinct letters can be selected from 26 or 5 distinct letters can be selected from 26 to make the 4 digit codes or 5 digit codes .

so if \("distinct "\) is not mentioned then we automatically should assume that there can be repetitions .

So in this question since no distinct word is mentioned , we can assume letters can we repeated to form the codes.Unlike the sum in the link above.

Hope this will prevent many people from wondering why we are solving two very similar questions in two very different ways. like I myself was wondering for a while before this eureka moment

if Anyone can add or verify or correct the reasoning that I have used It would certainly help.

There are two slightly different ways of doing this question:

No of different four letter codes:
26*26*26*26 = 26^4

No of different five letter codes:
26*26*26*26*26 = 26^5

Total number of codes = 26^4 + 26^5 = 27*26^4

or

Total no of different four or five letter codes:
26*26*26*26*27 = 27*26^4
(you multiply by 27 because for the last letter, you have 27 different possibilities - the 26 letters and null i.e. no letter which takes care of including all the 4 letter codes)

26^4+26^5 when we have "OR" word in sentence then when we add two posibilities and wen we have and word ..we multiple those posibilties
26^4(1+26)=27*26^4
so ans C..
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a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html

in the link posted above also contains a similar question of 4 letter code where A,B,C,A - two A's are repeating so we are using a formula 4 !/2 !
here also we are repeating the same letters tats why we are 26 ^4 for a letter code .But i should be 26 ^4 /4 ! na?

please help me i am getting confused..When should i use the principle n!/ no# repeating letters and when i should not?

Here each letter can come any number of times. i.e a 4 letter code can be aaaa.

But in the link provided by you, due to the restrictions imposed by the question, such liberty is not allowed there. There each letter should appear atleast once... leaving only 1 letter to repeat. Hence the difference.

Hope you understood it

In this case, wouldn't there be a possibility of 2 tickets having the same code? If no, can you please explain! Thanks

Which two codes could possibly be the same? It would be better to try with an easier example: try to count the number of 3 digit codes using 2 letters. You should get 2^3.

For more practice, check Constructing Numbers, Codes and Passwords in our Speciall Questions Directory.
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The first thing to note in these questions is whether we are allowed to repeat the variables or not.
Since here, nothing about repetition is mentioned, we can safely assume that we can repeat the variables.

4 Letter Code: _ _ _ _
The first place can have 26 alphabets.
The second place can also have 26 alphabets, since we can repeat.
Similarly for 3rd and 4th.
Hence total codes = 26*26*26*26 = \(26^4\)

5 Letter Code: _ _ _ _ _
By the above logic,
Total codes = \(26^5\)

Since we are asked the 4 letter codes OR the 5 letter codes,

Total codes = \(26^4 + 26^5\) = \(26^4(26 + 1)\) = \(26^4*27\)

Option C

1. The maximum is when alphabets are repeating and ordering is important, which is n^r
2. For 4 letter codes it is 26^4
3. for 5 letter codes it is 26^5
4. Total is (26^4 + 26^5)= 26^4(1+26) = 27(26)^4
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Can someone explain the step where we get (26 + 1). I follow up until that point. Unsure how we get that part.

That's done by factoring out 26^4 from both terms:

\(26^4+26^5=26^4+26^4*26=26^4(1+26)=27*26^4\).

Hope it's clear.
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We need to determine the maximum number of different stocks that can be designated by a 4-letter or 5-letter code that is created by using the 26 letters of the alphabet. Number of 5-letter codes:

26 x 26 x 26 x 26 x 26 = 26^5

Number of 4-letter codes:

26 x 26 x 26 x 26 = 26^4

Since the stocks can be designated by a 4-letter OR 5-letter code, we must add our results together to determine the maximum number of codes that can be created.

26^5 + 26^4 = 26^4(26 + 1) = 26^4(27)

Answer: C
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t

I don't understand the logic. If alphabets are repeating then there will be many similar codes.
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I used the 26c4 multipled 26c5 and got the answer wrong. Whats the reason for this and when should we use this formula?

26C4 gives the number of unordered groups of 4 different letters out of 26. For one, the order matters, {a, b, c, d} code is different from {b, a, c, d} code. Also, the letters could be repeated in the code, and 26C4 gives groups of 4 different letters. Finally, multiplying is wrong because 4-letter codes and 5-letter codes are different cases, so the number of possible codes should be added not multiplied.

Hope it's clear.
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thanks bossman. are these type of questions generally placed in the 700 level category or can be a product of the 600-700 level?

It depends. This one for example, is 600-700 level question. You can check different level combination's questions in our questions ban: https://gmatclub.com/forum/search.php?view=search_tags

Hope it helps.
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