# Permutation or binomial expansion?

• Feb 29th 2008, 10:44 PM
lllll
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?
• Mar 1st 2008, 12:25 AM
mr fantastic
Quote:

Originally Posted by lllll
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.