Advance Combination

[vineetbatra]

32. The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
(a) 246
(b) 248
(c) 492
(d) 15,128
(e) 30,256

Will post OA later

[NikolayZ]

hey Vineet!

I assume that the code can consist of 2 similar sings (AA, BB)

We know that 2 signs are unused.

1) 2 signs can be themselves 2!=2 codes, plus 2 (AA and BB), 4 overall.
2)
(a)With 2 unused signs we have 2 different first signs and 2 latter ones ( 2*2=4).
(b)We have another 122 signs to be the first or the second sign of the code. Combined with (a) we have 4*122 possibilities.
So, 122*4+4=492.

I think the answer is C.

[carllecat]

32. The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
(a) 246
(b) 248
(c) 492
(d) 15,128
(e) 30,256

Will post OA later

I don't know if my reasoning is good and maybe someone can tell...

According to me, this is a Permutation type problem, because the order is important. Therefore, you want to find the difference between P124 - P122, which means:

124 CODES: 124P/(124-2)P = 124*123 = 15252 possible permutations.

122 CODES: 122P/(122-2)P = 122*121 = 14762 possible permutations.

THE DIFFERENCE 124P - 122P = 15252 - 14762 = 490 (any reason why I am off by 2?)

My answer would be C

Any thoughts?

[Brent@GMATPrepNow]

32. The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
(a) 246
(b) 248
(c) 492
(d) 15,128
(e) 30,256
Will post OA later

I'll assume that we can use the same sign twice (since my answer matches one of the ones offered)

Each code consists of a first sign and a second sign.
With 124 signs, we have 124 options for the first sign and 124 options for the second sign for a total of 124^2 different codes.
With 122 signs, we have 122 options for the first sign and 122 options for the second sign for a total of 122^2 different codes.

The question essentially asks for the difference 124^2 - 122^2
The quick calculation here is to recognize that we have a difference of squares. So,
124^2 - 122^2 = (124+122)(124-122)
=246x2
= 492 = C

[vineetbatra]

All of you guys are correct the OA is infact C. Thanks for the response.

Vineet

[Scott@TargetTestPrep]

32. The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?
(a) 246
(b) 248
(c) 492
(d) 15,128
(e) 30,256

The 124 different signs can generate a total of 124 x 124 different 2-letter area codes. The 122 different signs can generate a total of 122 x 122 different 2-letter area codes. The difference between using 124 signs and 122 signs is:

124^2 - 122^2 = (124 - 122)(124 + 122) = 2 x 246 = 492

Answer: C