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 $879$

We have, $10$ white , $9$ green, $7$ black balls
$\therefore$ Total number of ways in which we can select balls as white $=\left(1W\right)+\left(2W\right)....+\left(10W\right)+(1$way to not select white one at all$)$
$=10+1=11$ ways in total for white ball.

Similarly for green, there are $9+1=10$ ways
and for black, there are $7+1=8$ ways.

there is one more case where we do not select any colour ball out of white, green and black, that is $1$ way.

But we have to select from these three colours only so we have to subtract this one way.

Thus total number of ways $=\left[\right(10+1\left)\right(9+1\left)\right(7+1\left)\right]-1$
$=879$