Question

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?

Answer 1

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.

Answer 2

then calculate like ZeHanz said, @msingh

Answer 3

@ZeHanz
is the answer 425