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Math Help - Combinatorics Word Problem! Answer CHECK

  1. #1
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    Combinatorics Word Problem! Answer CHECK

    Answer Check...

    27. b) Question: In how many ways can a president, vice-president, and treasurer be chosen from an organization with 12 members?

    My answer:

    C (12, 3) = 12! divided by 4! 3! = 3,326,400 different ways?


    28. a) Question: How many 4-digit #'s can be formed from the set A = {0,1,2,3,4,5,6} if there is no repetition? (**NOTE: 0123 is not a 4-digit number because it equals 123)

    My answer:
    You can choose the 4 digits from set A in (7 4) ways.
    # of digits = 7! 4! 3! x 4! -1
    = 840-1
    = 839

    * You subtract 1 to represent the 0123 4-digit number that would form (which doesn't count)

    b) Question: How many of the numbers in part a) are odd?

    My answer:
    Because we are selecting odd #'s only - there are just 5 choices for the last digit (1,3,5,7,9)

    Number of odd digits = 8 x 8 x 7 x 5
    = 2240 different odd numbers

    c) Question: How many of the numbers in part a) contain a 3?

    My answer: ?? DON't KNOW!!
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  2. #2
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    Answer Check...

    27. b) Question: In how many ways can a president, vice-president, and treasurer be chosen from an organization with 12 members?

    My answer:

    C (12, 3) = 12! divided by 4! 3! = 3,326,400 different ways?
    Your answer is a little large, is it not?. Try P(12,3)
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  3. #3
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    I am afraid all your answers are wrong ...
    27b) First of all C(12,3) = 12! / (9! 3!) = 220
    But anyway I don't think this is the right answer
    Suppose you have only 3 people
    Your calculation would lead to C(3,3)=1
    According to me there are 3! = 6 possibilities when there are 3 people (a,b,c)
    (a,b,c)=(P,VP,T)
    (a,b,c)=(P,T,VP)
    etc ...

    C(12,3) is the number of groups of 3 people among one group of 12 people
    For each group of 3, you must multiply by 3!=6

    28. a)There are 6 possibilities for the first digit (from 1 to 6, 0 being excluded)
    There are again 6 possibilities for the second digit (from 0 to 6 excluding the first digit already chosen)
    There are 5 possibilities for the third digit (from 0 to 6 excluding the 2 first digits already chosen)
    There are 4 possibilities for the fourth digit (from 0 to 6 excluding the 3 first digits already chosen)
    This makes 6x6x5x4=720

    b) Your answer (2240) is not consistent with your answer in question a (839)
    I mean you cannot have more odd 4-digits than the total number of 4-digits !
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  4. #4
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    Thanks for the reply.
    My new answers (after your response) have turned into..

    b) C(12,3) x 3! = 1320 different ways a president, vice president and treasurer can be chosen from the same organization.

    28. a) Would you subtract one from this answer considering that {0123} doesnt count since it creates 123. I Think thats what they were hinting at when they said that in the note. So it would be that answer - 1?

    b) Number of odd #s = 5 x 5 x 4 x 3 = 300 (Which is a lot less than 720 total digits)
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  5. #5
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    Never mind, ignore my thought for b) because you already accounted for the 0 being excluded.

    Thank you!
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