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Math Help - Combinatorics - HELP! ( FCP)

  1. #1
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    Combinatorics - HELP! ( FCP)

    Hi guys here is the question . An area code is the first 3 digits in a phone number and indicates the location of either the province or the city. In Canada, the following area codes exist:

    Manitoba 204
    Saskatchewan 306
    Québec (Québec City) 418
    Montreal 514
    Newfoundland 709
    Québec 450, 819
    British Columbia 250, 604, 778
    Alberta (south) 403
    Ontario 519, 613, 705, 807
    Yukon and NW Territories 867
    Toronto (Ontario) 289, 647
    Ontario (Toronto Metro) 416
    New Brunswick 506
    Alberta 780
    Nova Scotia 902

    Notice that there are 3 area codes for British Columbia: 250, 604 and 778. It will be necessary to
    add another area code as the population increases. The new area code cannot be the same as an
    existing code, it must begin with a 3 and end in an even number. Determine the number of
    possible area codes to choose from.

    A. 40
    B. 44
    C. 49
    D. 50


    What i attempted was to use the fundamental counting principle and see there is 3 spots first spot is taken so 3 _ _ second spot can be any one from 0-9 and the third one has to be even and cant be the same as any other area code meaning 4 possible choices 10*4 = 40 but the answer says 49 what did i do wrong?
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  2. #2
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    Re: Combinatorics - HELP! ( FCP)

    The new are code has to be 3 _ _

    There are ten ways to fill the 2nd digit (0 - 9).

    Ignoring the existing area codes, there are 5 ways to fill the last digit (0, 2, 4, 6, 8) *note that 0 is even

    By multiplication principle, there are 10 x 5 = 50 codes to choose from.

    However, the area code cannot be "306" as it is taken by Saskatchewan.

    Hence, the number of possible area codes to choose from = 50 - 1 = 49
    Thanks from Gurp925
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  3. #3
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    Re: Combinatorics - HELP! ( FCP)

    Understood thankyou!
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