Question

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from $10$ white, $9$ green and $7$ black balls is?

• A. $880$
• B. $629$
• C. $630$
• D. $879$

Answer 1

The correct option is D

$879$

Explanation for the correct option:

Step 1: According to the question given data.

Let $p$ be the number of white balls selected, $p=10$

Let $q$ be the number of green balls selected, $q=9$

Let $r$ be the number of black balls selected, $r=7$

Step 2: Finding number of ways of selection.

The total number of ways in which a selection of some or all of $(p+q+r)$,

where $p$ are alike, $q$ are alike, $r$ are alike is given by

$(p+1)(q+1)(r+1)$

Here, we need to choose at least one. So no empty case is allowed. Thus we need to subtract $1$ from it.

$$\begin{gathered} \Rightarrow (p+1)(q+1)(r+1)-1=(10+1)(9+1)(7+1)-1 \\ =\left(11\right)\left(10\right)\left(8\right)-1 \\ =880-1 \\ =879 \end{gathered}$$

Hence, option (D) $\mathbf{879}$ is the correct answer.