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All of the stocks on the over the counter market are [#permalink]
27 Jan 2012, 12:35

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71% (01:51) correct
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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

Re: Tricky combinatorics question [#permalink]
27 Jan 2012, 12:38

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Expert's post

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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;

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.

Re: All of the stocks on the over the counter market are [#permalink]
23 Jun 2012, 10:24

1

This post received KUDOS

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.. _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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? _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

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.

Re: All of the stocks on the over the counter market are [#permalink]
05 May 2013, 08:06

So you mean to say that if there is no restriction then we will use the method followed for this question but if there comes restrictions such as same letters should repeat twice or thrice we have to use n!/identical objects!

Is my understanding correct ? _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: All of the stocks on the over the counter market are [#permalink]
05 May 2013, 15:29

skamal7 wrote:

So you mean to say that if there is no restriction then we will use the method followed for this question but if there comes restrictions such as same letters should repeat twice or thrice we have to use n!/identical objects!

Is my understanding correct ?

Yep. Almost. The restriction in the question you provided is that only the given set of alphabets can be used. Another example of that can be like " How many 5 digit code words can be formed using the letters {A,A,B,B,B}" The answer to this is 5!/(2!*3!)

Re: All of the stocks on the over the counter market are [#permalink]
19 May 2014, 02:34

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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.

You hit on a topic that i was wondering about. Are you saying that we cannot use the 10C4 + 10C5 but we CAN use 10P4 + 10P5 because the permutation formula let's you account for repetitions?

Re: All of the stocks on the over the counter market are [#permalink]
25 Oct 2014, 10:44

Bunuel wrote:

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.

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

GMAT Prep 1 - 520 Oct 13, 2014 GMAT Prep 2 - 640 Oct 21, 2014 THE GMAT. Coming Soon. Oct 29, 2014.

Re: All of the stocks on the over the counter market are [#permalink]
26 Oct 2014, 06:07

Expert's post

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swanidhi wrote:

Bunuel wrote:

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.

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.

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