Question

A bag contains $4$ red, $3$ green and $5$ blue balls. In how many ways, $2$ red, $1$ green and $3$ blue balls can be drawn from the bag?

• A. $19$
• B. $18$
• C. $180$
• D. $60$

Answer 1

The correct option is C $180$

The balls available in the bag can be represented as:

We need to select $2$ red balls from $4$ red balls. This can be done in ${}^4{C}_2$ ways.

We need to select $1$ green ball from $3$ green balls. This can be done in ${}^3{C}_1$ ways.


We need to select $3$ blue balls from $5$ blue balls. This can be done in ${}^5{C}_3$ ways.

Selection of $2$ red balls from $4$ red balls, $1$ green ball from $3$ green balls and $3$ blue balls from $5$ green balls can be done in ${}^4{C}_2\times {}^3{C}_1\times {}^5{C}_3$ ways.

${}^4{C}_2\times {}^3{C}_1\times {}^5{C}_3=\frac{4!}{2!(4-2)!}\times \frac{3!}{1!(3-1)!}\times \frac{5!}{3!(5-3)!}$

${}^4{C}_2\times {}^3{C}_1\times {}^5{C}_3=\frac{4!}{2!2!}\times \frac{3!}{1!2!}\times \frac{5!}{3!2!}$

${}^4{C}_2\times {}^3{C}_1\times {}^5{C}_3=\left(6\right)\times \left(3\right)\times \left(10\right)$

$=180$

The number of ways in which $2$ red balls can be selected from $4$ red balls, $1$ green ball can be selected from $3$ green balls and $3$ blue balls can be selected from $5$ blue balls is $180$.

Therefore, option (c.) is the correct answer.