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# Each runner during a race is labeled with a unique one-letter code or

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Math Expert
Joined: 02 Sep 2009
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Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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30 Jun 2017, 11:22
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Difficulty:

35% (medium)

Question Stats:

73% (01:46) correct 27% (01:51) wrong based on 147 sessions

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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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

A. 325
B. 351
C. 377
D. 650
E. 676

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Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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30 Jun 2017, 11:35
1
2
Bunuel 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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

A. 325
B. 351
C. 377
D. 650
E. 676

No of label using one letter = 26

No of labels using two letters = 26*25 (first letter can be any 26 alphabet and second letter will be from the remaining 25 alphabet)

but this combination includes the reverse as well. for eg. AB & BA both are included. Hence to remove the reverse order, we need to divide by 2 i.e half the number of combinations

Therefore total number of unique codes = 26 + 26*25/2 = 351

Option B
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Re: Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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01 Jul 2017, 04:15
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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

Number of One Letter Codes $$= 26C1 = \frac{26!}{1! * 25!} = 26$$

Number of Two Letter Codes $$= 26C2 = \frac{26!}{2! * 24!}$$

Note as we repeatation of the digits is not allowed we will have to divide above equation by 2

$$= \frac{26!}{2! * 24! * 2}$$

$$= \frac{26 * 25}{2}$$

$$= 13 * 25 = 325$$

Total Possible Combinations $$= 26 + 325 = 351$$

Hence, Answer is B
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"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
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Posts: 188
Re: Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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15 Jul 2017, 00:05
ydmuley 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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

Number of One Letter Codes $$= 26C1 = \frac{26!}{1! * 25!} = 26$$

Number of Two Letter Codes $$= 26C2 = \frac{26!}{2! * 24!}$$

Note as we repeatation of the digits is not allowed we will have to divide above equation by 2

$$= \frac{26!}{2! * 24! * 2}$$

$$= \frac{26 * 25}{2}$$

$$= 13 * 25 = 325$$

Total Possible Combinations $$= 26 + 325 = 351$$

Hence, Answer is B

Can you explain this with example?
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Re: Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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15 Jul 2017, 01:47
rocko911 wrote:
ydmuley 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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

Number of One Letter Codes $$= 26C1 = \frac{26!}{1! * 25!} = 26$$

Number of Two Letter Codes $$= 26C2 = \frac{26!}{2! * 24!}$$

Note as we repeatation of the digits is not allowed we will have to divide above equation by 2

$$= \frac{26!}{2! * 24! * 2}$$

$$= \frac{26 * 25}{2}$$

$$= 13 * 25 = 325$$

Total Possible Combinations $$= 26 + 325 = 351$$

Hence, Answer is B

Can you explain this with example?

Not sure what do you mean by example here? For these type of questions I always prefer to go by what the question is asking for.
_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
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Re: Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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15 Jul 2017, 05:03
ydmuley wrote:
rocko911 wrote:
ydmuley 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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

Number of One Letter Codes $$= 26C1 = \frac{26!}{1! * 25!} = 26$$

Number of Two Letter Codes $$= 26C2 = \frac{26!}{2! * 24!}$$

Note as we repeatation of the digits is not allowed we will have to divide above equation by 2

$$= \frac{26!}{2! * 24! * 2}$$

$$= \frac{26 * 25}{2}$$

$$= 13 * 25 = 325$$

Total Possible Combinations $$= 26 + 325 = 351$$

Hence, Answer is B

Can you explain this with example?

Not sure what do you mean by example here? For these type of questions I always prefer to go by what the question is asking for.

I meant that this was my first time to see a question like that... I solved it almost but how can I understand in just 2 minutes that i need to divide by 2 as well , what made u think that? it took me time to analyze how to do it
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Re: Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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15 Jul 2017, 08:10
rocko911 wrote:
ydmuley wrote:

Can you explain this with example?

Not sure what do you mean by example here? For these type of questions, I always prefer to go by what the question is asking for.

I meant that this was my first time to see a question like that... I solved it almost but how can I understand in just 2 minutes that I need to divide by 2 as well, what made u think that? it took me time to analyze how to do it[/quote]

I divided by 2 based on the comment - if a given two letter code is used then the reverse code is not used.

For these type of questions, leveraging the given information is very important (actually this applies to all the Quant questions), but in these type of problems you can lay down a structure very clearly.
_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
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Re: Each runner during a race is labeled with a unique one-letter code or  [#permalink]

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24 Jul 2017, 21:25
1
ydmuley 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, and if a given two letter code is used then the reverse code is not used. What is the maximum number of runners that can receive unique codes for the race?

Number of One Letter Codes $$= 26C1 = \frac{26!}{1! * 25!} = 26$$

Number of Two Letter Codes $$= 26C2 = \frac{26!}{2! * 24!}$$

Note as we repeatation of the digits is not allowed we will have to divide above equation by 2

$$= \frac{26!}{2! * 24! * 2}$$

$$= \frac{26 * 25}{2}$$

$$= 13 * 25 = 325$$

Total Possible Combinations $$= 26 + 325 = 351$$

Hence, Answer is B

I don't think you need to divide by that extra 2... it's already done in the combination formula. Isn't what you've written = (26*25)/4?
Re: Each runner during a race is labeled with a unique one-letter code or   [#permalink] 24 Jul 2017, 21:25
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