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by bryan88 » Sat Apr 14, 2012 4:12 am
A certain stock exchange designates each stock with a one-, or 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 with these codes?

A. 2,951
B. 8,125
C. 15,600
D. 16,302
E. 18,278
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by sam2304 » Sat Apr 14, 2012 4:22 am
Is it E ?

one letter code - 26
two letter code - 26 * 26
three letter code - 26^3

Sum of all the three = 18278.

Find the combinations and 6 raised to any power will give 6 as the units digit. Adding 3 sixes will yield 8 as the unit's digit. Don't try to calculate 26^3 and mess it up. I found this way easier :)
Last edited by sam2304 on Sat Apr 14, 2012 4:54 am, edited 1 time in total.
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by killer1387 » Sat Apr 14, 2012 4:26 am
bryan88 wrote:A certain stock exchange designates each stock with a one-, or 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 with these codes?

A. 2,951
B. 8,125
C. 15,600
D. 16,302
E. 18,278
one digit= 26

two digit ;
same digit= 26
different one's = 26*25 = 650

three digit;
same digit =26
all different = 26*25*24 = 15600
two same = 26* 3* 25 = 1950

adding all above = 18278

Hence E
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by bryan88 » Sat Apr 14, 2012 4:43 am
What differentiates this from a normal 26C1*26C2*26C3 problem?
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by sam2304 » Sat Apr 14, 2012 4:49 am
bryan88 wrote:What differentiates this from a normal 26C1*26C2*26C3 problem?
That is without repetition whereas it is given as the repetition is allowed. Moreover there is no need for multiplication in 26C1 * 26C2 * 26C3 - multiplication is used when you are to combine all the three as in for a 6 digit code but we are calculating independently for one, two and three digit codes so use addition.
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by Shalabh's Quants » Sat Apr 14, 2012 4:52 am
bryan88 wrote:A certain stock exchange designates each stock with a one-, or 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 with these codes?

A. 2,951
B. 8,125
C. 15,600
D. 16,302
E. 18,278
1 Digit Code...

=> 26.

2 Digits code...

=> Options available to fill 1st place=26;
Options available to fill 2nd place=26; As letters can be used twice/thrice. Repetition allowed.

Total ways = 26*26 = 676.

3 Digits code...

=> Options available to fill 1st place=26;
Options available to fill 2nd place=26; As letters can be used twice/thrice. Repetition allowed.
Options available to fill 3rd place=26; As letters can be used twice/thrice. Repetition allowed.


Total ways = 26*26*26 = 17,576.

Total code = 26 + 676 + 17576 = 18278.
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by penguinfoot » Sun Apr 15, 2012 4:51 am
bryan88 wrote:What differentiates this from a normal 26C1*26C2*26C3 problem?
Combination is used when you are picking some (may also be 1) items from a set of items !
The problem on the other hand, does not require you to pick any set of alphabets from A-Z, Moreover, it states repetition is allowed.

we have _ codes, _ _ codes, _ _ _ codes. Each blank can be filled by one of 26 alphabets.
Solution : 26 + 26*26 + 26*26*26 : [spoiler]Option (E)[/spoiler]

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