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# A box contains 4 red,3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colors?

Hint:The combination is the making of a group without considering the arrangement or order. Here a random selection of balls is needed. So, we need to compute the combination of different possibilities with further multiplication.

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.

So, the correct answer is “Option D”.

Additional Information:Combination is the field of mathematics concerned with problems of selection and arrangement. It involves the various operations within a finite or discrete system. It also includes the closely related area of combinatorial geometry.

Note:In these types of problems selection of the objects from a box in random order are considered as independent of previous selection. It will not depend on the fact that whether these are being done with replacement or without replacement.