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Math Help - Numbers & Operations Q from SAT Sbj Math II Test Study

  1. #1
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    Numbers & Operations Q from SAT Sbj Math II Test Study

    I'm sorry if this is in the wrong sub-forum. I'm guessing this could be classified as pretty basic stat. But here's the problem:

    "From a group of 6 juniors and 8 seniors on the student council, 2 juniors and 4 seniors will be chosen to make up a 6-person committee. How many different 6-person committees are possible?

    a. 84
    b. 85
    c. 1050
    d. 1710
    e. 1890
    "

    Then it tells me:
    "Choice (c) is the correct answer. The 2 juniors on the committee can be chosen from the 6 juniors in (6/2) = 15 ways. The 4 seniors on the committee can be chosen from the 8 seniors in (8/4) = 70 ways. Therefore there are (15)(70) = 1050 possibilities for the 6 person committee."

    I'm pretty bad at these.. Can someone please explain (6/2) = 15 and (8/4) = 70 deal to me? Or maybe just show me a clearer way of doing this? Thanks!
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  2. #2
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    Hello, Savior_Self!

    rom a group of 6 juniors and 8 seniors on the student council,
    2 juniors and 4 seniors will be chosen to make up a 6-person committee.
    How many different 6-person committees are possible?

    . . (a)\;84 \qquad (b)\; 85 \qquad (c)\;1050 \qquad (d)\; 1710 \qquad (e)\; 1890

    Then it tells me: ."Choice (c) is the correct answer.
    The 2 juniors on the committee can be chosen from the 6 juniors in (6/2) = 15 ways.
    The 4 seniors on the committee can be chosen from the 8 seniors in (8/4) = 70 ways.
    Therefore, there are (15)(70) = 1050 possibilities for the 6 person committee."

    Can someone please explain (6/2) = 15 and (8/4) = 70 deal to me?

    Those expressions are not fractions; they are "combination" numbers.


    Selecting 2 juniors from 6 juniors, there are: \displaystyle {6\choose2} \:=\:\frac{6!}{2!\,4!} \:=\:15 ways.

    Selecting 4 seniors from 8 seniors, there are: \displaystyle {8\choose4} = \frac{8!}{4!\,4!} \:=\:70 ways.


    Maybe you'd recognize them if they were written like this?

    . . . . . _6C_2\:\text{ and }\:_8C_4
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