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bryan88 Really wants to Beat The GMAT! Default Avatar
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GMAT prep! Post Sat Apr 14, 2012 5:12 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    sam2304 GMAT Titan
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    Post Sat Apr 14, 2012 5:22 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 Smile

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    Last edited by sam2304 on Sat Apr 14, 2012 5:54 am; edited 1 time in total

    killer1387 GMAT Destroyer! Default Avatar
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    Post Sat Apr 14, 2012 5:26 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

    bryan88 Really wants to Beat The GMAT! Default Avatar
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    Post Sat Apr 14, 2012 5:43 am
    What differentiates this from a normal 26C1*26C2*26C3 problem?

    sam2304 GMAT Titan
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    Post Sat Apr 14, 2012 5:49 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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    Shalabh's Quants Really wants to Beat The GMAT!
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    Post Sat Apr 14, 2012 5:52 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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    penguinfoot Really wants to Beat The GMAT! Default Avatar
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    Post Sun Apr 15, 2012 5:51 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 : Option (E)

    Penguinfoot.

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