Questions & Answers

Question

Answers

Answer
Verified

Here, 4 red balls, 3 white balls and 2 blue balls are in the box. And 3 balls are drawn at random. This is a simple situation. We will compute the possible ways one by one for each color.

Case-1: 1 red ball can be selected in ${}^4{C_1}$ ways.

Case-2: 1 white ball can be selected in ${}^3{C_1}$ ways.

Case-3: 1 blue ball can be selected in ${}^2{C_1}$ ways.

All three cases will occur to get 3 balls with required color.

Therefore, the total number of ways for having total 3 balls each of different colors.

$= {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1}$

$= \dfrac{{4!}}{{1! \times 3!}} \times \dfrac{{3!}}{{1! \times 2!}} \times \dfrac{{2!}}{{1! \times 1!}}$

$= 4 \times 3 \times 2$

$= 24$

Here, we have multiplied the combination results to get the final one. This is because we have to multiply all possibilities when there is AND condition. In this question it is the AND operation.

Thus the total number of ways will be 24.