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Math Help - Combinatorics

  1. #1
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    Combinatorics

    27. a) In how many ways can a committee of 3 be chosen from an organization with 12 members?

    My answer: Number of Ways (12 3) = 220 ways

    b) In how many ways can a president, vice-president, and treasurer be chosen from this same organization?

    My answer: C (12, 3) = 12! divided by 4! 3! = 3,326,400 different ways


    28. a) How many 4-digit numbers 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 # because it equals 123)

    My answer: Number of digits = (7 4) x 4! -1 ( minus one to get rid of the 0123 choice)
    = 839 4 digit numbers can be formed

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

    My answer: number of odd digits = 8 x 8 x 7 x 5 = 2240 odd digits This doesn't make sense because there can't be more odd digits than there are total choices in a)

    b) How many of the numbers in part a) contain a 3?

    ** any clues?

    Ahh, this is due tomorrow so please let me know if they are right! Thank you
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  2. #2
    Senior Member
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    27. a) In how many ways can a committee of 3 be chosen from an organization with 12 members?

    My answer: Number of Ways (12 3) = 220 ways
    Correct

    b) In how many ways can a president, vice-president, and treasurer be chosen from this same organization?

    My answer: C (12, 3) = 12! divided by 4! 3! = 3,326,400 different ways
    Not sure how you got this. There are 12 choices for president, 11 for vice and 10 for secretary

    28. a) How many 4-digit numbers 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 # because it equals 123)

    My answer: Number of digits = (7 4) x 4! -1 ( minus one to get rid of the 0123 choice)
    = 839 4 digit numbers can be formed
    I think the point of the 0123 example was to say that no numbers with a 0 at the front would be considered as 4 digit numbers.


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

    My answer: number of odd digits = 8 x 8 x 7 x 5 = 2240 odd digits This doesn't make sense because there can't be more odd digits than there are total choices in a)
    Not sure what you are doing here. The 4-digit number will be odd if and only if the last digit is odd. How many of the 4-digit numbers will have an odd last digit?


    b) How many of the numbers in part a) contain a 3?
    Find the number of 3-digit numbers you can make from the set {0,1,2,4,5,6}. Then multiply by the number of places you can add a 3 to those numbers. You will need to be careful with things like 3012 and 0312.
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