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# Each runner during a race is labeled with a unique

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Each runner during a race is labeled with a unique [#permalink]

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16 Sep 2004, 20:54
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Each runner during a race is labeled with a unique one-letter code or a unique two-letter code, where the two letters are different. The codes use the 26-letter English alphabet. What is the maximum number of runners that can receive unique codes for the race?
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16 Sep 2004, 21:03
676
26 + 26^2 - 26 = 26^2 = 676
First 26 is just the 26 individual letters
26^2 is for all possible combinations of 2 alphabet letters
Last 26 is unfavorable outcomes when 2 letters are similar ie AA,BB,CC... There are 26 of these possibilities
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16 Sep 2004, 22:10
First letter can be filled in 26 ways.
Second letter can be filled in 26 ways (25 alphabets or 1 space).
Hence 26 x 26 = 676
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17 Sep 2004, 00:21
Each runner during a race is labeled with a unique one-letter code or a unique two-letter code, where the two letters are different. The codes use the 26-letter English alphabet. What is the maximum number of runners that can receive unique codes for the race?

26 unique one-letter codes
(26*25) unique two letter codes where two letters are different = 650

SO 650+26 = 676
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17 Oct 2011, 08:48
26+(26*26-26)/2=351
division by 2 to make sure we don't count 2 letter code twice i.e. AB & BA should be count just once.

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17 Oct 2011, 08:56
total no. of alphabets = 26
pattern of code: 12
At place 1 we can have null or 26 characters.
At place 2 we can have 26 characters or 25 characters(depending if null was at 1 or not)
total becomes 1*26 + 26*25 = 26+ 650 = 676

or faster way is to start from right side
at place 2, we have 26 characters
at place 1 we have 25 character +1 null= 26
26*26=676
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18 Oct 2011, 04:32
Unique letter combinations are 26
two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325
two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

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18 Oct 2011, 07:23
Expert's post
nishtil wrote:
Unique letter combinations are 26
two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325
two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

Would you say AC and CA are different codes or same?
I think you will agree that they are different codes. Hence you do not use the combination formula here (26C2) because it only selects 2 different letters out of the 26 letters.

So how do you solve this question? You can do it in two different ways:

Method 1:
You select 2 different letters and then arrange them.
26C2 * 2! = 650

Method 2:
You use basic counting principle. (Check out: http://www.veritasprep.com/blog/2011/10 ... inatorics/)
26*25 = 650

The number of one letter codes is 26 so total = 650 + 26 = 676
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19 Oct 2011, 00:37
VeritasPrepKarishma wrote:
nishtil wrote:
Unique letter combinations are 26
two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325
two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

Would you say AC and CA are different codes or same?
I think you will agree that they are different codes. Hence you do not use the combination formula here (26C2) because it only selects 2 different letters out of the 26 letters.

So how do you solve this question? You can do it in two different ways:

Method 1:
You select 2 different letters and then arrange them.
26C2 * 2! = 650

Method 2:
You use basic counting principle. (Check out: http://www.veritasprep.com/blog/2011/10 ... inatorics/)
26*25 = 650

The number of one letter codes is 26 so total = 650 + 26 = 676

I agree with you and understood your point. however in one of kaplan CAT the OA is porvided as 351. Please see the attachment. Please anyone explain why OA is 351
Attachment:

Kaplan CAT.PNG [ 68.02 KiB | Viewed 1349 times ]

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19 Oct 2011, 01:53
nishtil wrote:
VeritasPrepKarishma wrote:
nishtil wrote:
Unique letter combinations are 26
two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325
two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

Would you say AC and CA are different codes or same?
I think you will agree that they are different codes. Hence you do not use the combination formula here (26C2) because it only selects 2 different letters out of the 26 letters.

So how do you solve this question? You can do it in two different ways:

Method 1:
You select 2 different letters and then arrange them.
26C2 * 2! = 650

Method 2:
You use basic counting principle. (Check out: http://www.veritasprep.com/blog/2011/10 ... inatorics/)
26*25 = 650

The number of one letter codes is 26 so total = 650 + 26 = 676

I agree with you and understood your point. however in one of kaplan CAT the OA is porvided as 351. Please see the attachment. Please anyone explain why OA is 351
Attachment:
Kaplan CAT.PNG

Yes, because the question in the attachment is different from the main post. It states an additional condition that if a 2 letter code is used, the reverse cannot be used.

So,
AB- Is one valid code
BA- will not be a valid code because AB has already been used.

Whereas according to the main question, both AB and BA will be considered valid codes.

Original post's Answer: $$P^{26}_{1}+P^{26}_{2}$$

Attachment's Answer: $$C^{26}_{1}+C^{26}_{2}$$
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19 Oct 2011, 02:06
Thanks fluke. This is the precise reason I opened a saparate topic as combinatorics-kaplan-700-level-ps-unique-2-latter-code-122067.html. This was locked by the 144144. can you please open the thread so that discussion can go forward
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19 Oct 2011, 20:04
Bhai wrote:
Each runner during a race is labeled with a unique one-letter code or a unique two-letter code, where the two letters are different. The codes use the 26-letter English alphabet. What is the maximum number of runners that can receive unique codes for the race?

Ans- 676

Use slot method to solve this question:

For unique one letter code: 26
For unique two letter code: 26*25 = 650

Total # of codes = 26 + 650 = 676
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Re: Combimation: Letters   [#permalink] 19 Oct 2011, 20:04
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