Question

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Answer 1

Hint: In this problem the bag contains 5 black and 6 red balls. We have to find the number of ways in which 2 black and 3 red balls can be selected, one thing is for sure that 2 black balls can only be drawn from the overall 5 black balls extending this concept to the red balls, 3 red balls can be selected from overall 6 red balls only. Use this concept to reach the solution.

The bag has 5 black balls and 6 red balls.
Now the number of ways of selecting 2 black balls from in total 5 black balls will be ${}^5{C}_2$…………… (1)
Now the number of ways of selecting 3 red balls from in total 6 red balls will be ${}^6{C}_3$…………… (2)
The total number of ways of selecting 2 black balls and 3 red balls will be equation (1) multiplied with equation (2).
${\Rightarrow }^5{C}_2{\times }^6{C}_3$…………….. (3)
Using the formula of ${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$ we can rewrite equation (3) as
$$\begin{gathered} {\Rightarrow }^5{C}_2{\times }^6{C}_3={\displaystyle \frac{5!}{2!\left(5-2\right)!}}\times {\displaystyle \frac{6!}{3!\left(6-3\right)!}} \\ \Rightarrow {\displaystyle \frac{5!}{2!\left(3\right)!}}\times {\displaystyle \frac{6!}{3!\left(3\right)!}} \end{gathered}$$
Using the concept that $n!=n\times (n-1)(n-2)(n-3).......(n-r)!\text{ where r < n}$.
$$\begin{gathered} {}^5{C}_2{\times }^6{C}_3={\displaystyle \frac{5\times 4\times 3!}{2\left(3\right)!}}\times {\displaystyle \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times \left(3\right)!}} \\ \Rightarrow 10\times 20=200 \end{gathered}$$
The number of ways in which 2 black and 3 red balls can be selected is 200.
Note: Whenever we face such types of problems the key concept is to have the physical understanding of the formula ${}^n{C}_r$ which is the number of ways of selecting any r entities out of n entities. This concept along with the mathematical formula will help you solve problems of this kind and will take you to the right answer.