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 $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ we can rewrite equation (3) as
$
  { \Rightarrow ^5}{C_2}{ \times ^6}{C_3} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} \times \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} \\
   \Rightarrow \dfrac{{5!}}{{2!\left( 3 \right)!}} \times \dfrac{{6!}}{{3!\left( 3 \right)!}} \\
$
Using the concept that $n! = n \times (n - 1)(n - 2)(n - 3).......(n - r)!{\text{ where r < n}}$.
$
  ^5{C_2}{ \times ^6}{C_3} = \dfrac{{5 \times 4 \times 3!}}{{2\left( 3 \right)!}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3 \times 2 \times 1 \times \left( 3 \right)!}} \\
   \Rightarrow 10 \times 20 = 200 \\
$
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.