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Permutation and Combination Problem, Need help

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by gmattester » Wed Jul 30, 2008 11:54 am
A certain stock exchange designates each stock with a one-, two-, or three letter code, where each letter is selected from the 26 letters of the alphabet. if the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate these codes?

a) 2951
b) 8125
c) 15600
d) 16302
e) 18278

Can someone please explain how to solve this question.
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Source: — Problem Solving |

by swati.sug » Wed Jul 30, 2008 12:01 pm
If code is one letter
No of possible codes = 26

If code is two letters
No of possible codes = 26*26 =676

If code is three letters
No of possible codes = 26*26*26 =17576

Total = 26+676 +17576 =18278

Hope that helps!
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by parallel_chase » Wed Jul 30, 2008 12:08 pm
Even I got 18278

26+676 +17576 =18278

Whats the OA?
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Can you explain this

by gmattester » Wed Jul 30, 2008 12:12 pm
Thanks swati......
Answer is right.

Can you explain this. If we are looking for different stocks possible to uniquely designate these codes then

If code is one letter
No of possible codes should be 26

If code is two letters
No of possible codes should be 26*25= 650

If code is three letters
No of possible codes should be 26*25*24=15600

Total = 26+650 +15600
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by ricky » Wed Jul 30, 2008 12:51 pm
Since we can repeat a letter in the code, therefore for a two letter code the number of possibilities are 26* 26 and NOT 26*25.Same is true for a three letter code.
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