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All of the stocks on the over the counter market are [#permalink]
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27 Jan 2012, 13:35
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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
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Re: Tricky combinatorics question [#permalink]
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27 Jan 2012, 13:38
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Re: All of the stocks on the over the counter market are [#permalink]
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23 Jun 2012, 03:30
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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 ifacodewordisdefinedtobeasequenceofdifferent126652.htmlIn 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.



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Re: All of the stocks on the over the counter market are [#permalink]
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23 Jun 2012, 11:24
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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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Re: All of the stocks on the over the counter market are [#permalink]
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05 May 2013, 07:14
a4lettercodewordconsistsoflettersabandcifthe59065.htmlin 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?
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Re: All of the stocks on the over the counter market are [#permalink]
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05 May 2013, 09:00
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skamal7 wrote: http://gmatclub.com/forum/a4lettercodewordconsistsoflettersabandcifthe59065.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



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Re: All of the stocks on the over the counter market are [#permalink]
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05 May 2013, 09:06
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 ?
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Re: All of the stocks on the over the counter market are [#permalink]
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05 May 2013, 16:29
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!) Hope you got your answer.



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Re: All of the stocks on the over the counter market are [#permalink]
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19 May 2014, 03:34
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Re: All of the stocks on the over the counter market are [#permalink]
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23 Aug 2014, 10:43
Joy111 wrote: 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 ifacodewordisdefinedtobeasequenceofdifferent126652.htmlIn 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. 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?



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Re: All of the stocks on the over the counter market are [#permalink]
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25 Oct 2014, 11:44
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 4digit code {XXXX} each digit can take 26 values (as there are 26 letters), so total # of 4digits code possible is 26^4;
The same for 5digit code {XXXXX} again each digit can take 26 values (26 letters), so total # of 5digits 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
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Re: All of the stocks on the over the counter market are [#permalink]
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26 Oct 2014, 07:07
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 4digit code {XXXX} each digit can take 26 values (as there are 26 letters), so total # of 4digits code possible is 26^4;
The same for 5digit code {XXXXX} again each digit can take 26 values (26 letters), so total # of 5digits 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. For more practice, check Constructing Numbers, Codes and Passwords in our Speciall Questions Directory.
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Re: All of the stocks on the over the counter market are [#permalink]
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Re: All of the stocks on the over the counter market are [#permalink]
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09 Nov 2015, 00:41
Murmeltier 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 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
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Re: All of the stocks on the over the counter market are [#permalink]
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