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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?

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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ydmuley
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.
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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?

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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ydmuley



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.
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ydmuley
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?
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"Unique two-letter code" = slots or permutations

"Reverse of code cannot be used" = half of the results don't count so divide permutations result by 2

"Or" = add
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