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19 June, 13:02

A club with six members is to choose three officers: president, vice-president, and secretary-treasurer. If each office is to be held by one person and no person can hold more than one office, in how many ways can those offices be filled?

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  1. 19 June, 14:15
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    120 ways

    Step-by-step explanation:

    Total numbers of members 'n' = 6

    number of officers to choose 'r' = 3

    So, for the number of permutation of (n=6) members taken (r=3) at a time, the formula is

    nPr = n! / (n-r) !

    6P3 = 6! / (6-3) !

    6P3 = 6! / 3! = > (6x 5x4x3x2x1) / (3x2x1)

    6P3 = 120

    Thus, those officers can be filled in 120 ways
  2. 19 June, 14:25
    0
    120 ways

    Step-by-step explanation:

    We have a total of 6 people, and we want to form groups of 3, so we can do a combination of 6 choose 3.

    But inside a group of 3 people, we need to choose the office of each one, and the number of possibilities for this is calculated using factorial.

    So, we have the following:

    Number of groups of 3 among 6 people:

    C (6,3) = 6! / (3!*3!) = 6*5*4/6 = 20

    Different offices inside the group of 3:

    3! = 3*2 = 6

    Then, to find the total number of possibilities that offices can be filled, we multiply these results:

    20 * 6 = 120

    (This problem can also be solved using permutation, as the order of the elements in the group matters:

    P (6,3) = 6!/3! = 6*5*4 = 120)
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