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Math Help - Permutation or binomial expansion?

  1. #1
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    Permutation or binomial expansion?

    A group of 3 undergraduates and 5 graduate students are available to fill a position. If 4 students are to be randomly selected from this group find the probability that exactly 2 undergraduates will be among the chosen 4.

    I'm thinking that I have to resolve this problem using a binomial expansion with \begin{pmatrix} 8 \\ 4 \end{pmatrix} which gives me \frac{8!}{4!(4!)} possible ways of filling the post and then \begin{pmatrix} 3 \\ 2 \end{pmatrix} to find the number of undergrads and \begin{pmatrix} 5 \\ 2 \end{pmatrix} for the number of grads. which would give me a final solution of

    \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix}}{\begin{pmatrix} 8 \\ 4 \end{pmatrix}} = \frac{(\frac{3!}{2!(1!)})(\frac{5!}{2!(3!)})}{\fra  c{8!}{4!(4!)}} = \frac{(3)(10)}{70} = \frac{3}{7}

    Is this right?
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  2. #2
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    Quote Originally Posted by lllll View Post
    A group of 3 undergraduates and 5 graduate students are available to fill a position. If 4 students are to be randomly selected from this group find the probability that exactly 2 undergraduates will be among the chosen 4.

    I'm thinking that I have to resolve this problem using a binomial expansion with \begin{pmatrix} 8 \\ 4 \end{pmatrix} which gives me \frac{8!}{4!(4!)} possible ways of filling the post and then \begin{pmatrix} 3 \\ 2 \end{pmatrix} to find the number of undergrads and \begin{pmatrix} 5 \\ 2 \end{pmatrix} for the number of grads. which would give me a final solution of

    \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix}}{\begin{pmatrix} 8 \\ 4 \end{pmatrix}} = \frac{(\frac{3!}{2!(1!)})(\frac{5!}{2!(3!)})}{\fra  c{8!}{4!(4!)}} = \frac{(3)(10)}{70} = \frac{3}{7}

    Is this right?
    Yes.
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