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Permutations Problem

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by sreak1089 » Wed Jan 20, 2010 12:18 am
A telephone company wanted 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) 15128
E) 30256
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by fibbonnaci » Wed Jan 20, 2010 12:53 am
each letter can take 122 different signs.
total ways of forming 2 letters is 122*122

Now if the company wants to use 124 signs, then for 2 letters ,
total number of ways = 124*124.

Difference= (124)^2 - (122)^2
= (124+ 122) (124-122)
= 246*2 => 492 additional area codes can be created.

Answer = C
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by sreak1089 » Wed Jan 20, 2010 7:55 am
I am not able to figure out, how you got 122*122? can you please elaborate?

Is it because, you have 122 signs for the first letter and for each of the 122 letters you can have further 122 signs right?

To make it simple, let's say you have only 3 symbols (assume a, b, c) if you have two letter code then you have:

aa ba ca
ab bb cb
ac bc cc

Thus, 3 * 3 = 9 ways.

I thinking I am getting it, Am I correct?[/img]
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by fibbonnaci » Wed Jan 20, 2010 7:58 am
exactly.
for the first letter- you have 3 ways.(a, b, c)
and for the next letter similiar 3 ways.
so total 3*3 ways.

Hope i have clarified ur doubt!
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by sreak1089 » Wed Jan 20, 2010 8:17 am
Yes, sir you have. Thank you so much.
fibbonnaci wrote:exactly.
for the first letter- you have 3 ways.(a, b, c)
and for the next letter similiar 3 ways.
so total 3*3 ways.

Hope i have clarified ur doubt!
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