OpenStudy (anonymous):

A box contains 5 red balls and 6 black balls. in how many ways can 6 balls be selected so that there are at least two balls of each color?

5 years ago
OpenStudy (zehanz):

So you want: 1) 2 red, 4 black or 2) 3 red, 3 black or 3) 4 red, 2 black 1): (number of ways to pick 2 out of 5) times (noof ways to pick 4 out of 6): $\left(\begin{matrix}5 \\ 2\end{matrix}\right) \cdot \left(\begin{matrix}6 \\ 4\end{matrix}\right)$ 2): $\left(\begin{matrix}5 \\ 3\end{matrix}\right) \cdot \left(\begin{matrix}6 \\ 3\end{matrix}\right)$ 3):$\left(\begin{matrix}5 \\ 4\end{matrix}\right) \cdot \left(\begin{matrix}6 \\ 2\end{matrix}\right)$ You can calculate these binomial coefficients by the following rule:$\left(\begin{matrix}n \\ k\end{matrix}\right)= \frac{n!}{k!(n-k)!}$So$\left(\begin{matrix}5 \\ 3\end{matrix}\right)= \frac{5!}{2!3!}=\frac{4 \cdot 5}{2}=10$ Or use a calculator with the nCr function. Add up the outcomes of 1), 2) and 3) to get the final answer.

5 years ago

then calculate like ZeHanz said, @msingh

5 years ago
OpenStudy (anonymous):