If the reading club has nine members, how many ways can the club choose five members to attend a special program if order does not matter?

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  • David
    Lv 7
    1 month ago

    Any 5 from 9 = 126 using your calculator

  • 1 month ago

    9C5 =

    9!/(5!(9 - 5)!) =

    126

  • 1 month ago

    Answer  =   126

    It is case of finding combinations of 5 out of 9 members ( since order does not matter).  Formula used is -------

    ..   ...  .    ...n !

    ⁿCᵣ = -------------------  .  In this case  n = 9  and  r = 5

    ...    .. r !  *  (n-r) !

    .................    9!  ................9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

    ie  ⁹C₅  = ---------------  =  -------------------------------------------  =  126

    ...       ...    5! * (9-5)!     .... 5 * 4 * 3 * 2 * 1  x  4 * 3 * 2 * 1

    Answer  =   126

  • 1 month ago

    thats up to them who they choose

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  • 1 month ago

    The answer is "9 choose 5"

    To calculate that, start by writing 9/5 as a fraction:

    9

    --

    5

    Now multiply by the next smaller number on top and bottom. Repeat until you get to 1 on the bottom.

    9 * 8 * 7 * 6 * 5

    --------------------

    5 * 4 * 3 * 2 * 1

    If you are doing this by hand, you can cancel everything on the bottom. For example, you can cancel 5 and 5. You can also cancel 8 and 4 * 2. Finally you can partially cancel 9 and 3.

    3 * 1 * 7 * 6 * 1

    --------------------

    1 * 1 * 1 * 1 * 1

    Get rid of all the ones.

    3 * 7 * 6

    = 126 ways

    As a double-check, go to Google and type "9 choose 5"

  • alex
    Lv 7
    1 month ago

    9C5= .... (your calculator)

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