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Kaplan Test - Codes

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by singhsa » Sat Sep 25, 2010 11:26 pm
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. 677
d. 650
e. 676

OA - B
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by narik11 » Sat Sep 25, 2010 11:59 pm
singhsa 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. 677
d. 650
e. 676

OA - B
Case 1: Single letter : 26

Case 2: Two letters: Summation of 26 = (26*27)/2 =351

I derived this taking A,B,C,D,E

so i had..

AA BB CC DD EE
AB BC CD DE
AC BD CE
AD BE
AE

Since u have two letters to be different u will not have AA,BB,CC,DD,EE

So therefore 26+351-26=351
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by Gurpinder » Sun Sep 26, 2010 6:57 am
singhsa 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. 677
d. 650
e. 676

OA - B
first take the 26 alone.

26!/2!24! = 325

325+26=351

hence (B)
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by Stuart@KaplanGMAT » Sun Sep 26, 2010 8:24 am
singhsa 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. 677
d. 650
e. 676

OA - B
Another way you can think about it:

one letter codes: 26 possibilities

two letter codes: 26 possibilities for the first letter, leaving 25 possibilities for the second letter.

26*25 = 650

However, we've ignored the rule about not reusing the same two letters (under our method, we've counted AB and BA as two separate codes). So, we need to divide by 2 to get rid of the duplicates:

650/2 = 325

Total: 26 + 325 = 351
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