Question

Out of $10$ white, $9$ black and $7$ red balls, the number of ways in which selection of one or more balls can be made, is:
A) $881$
B) $891$
C) $879$
D) $892$

Answer 1

Hint: You have to choose one or more balls from the three boxes of different colored balls respectively. So you either choose ball from a box or do not choose, and choose from other or vice versa, so in this way you will have possible number of way to select ball from a box is one more than the total number of balls present in it and in order to find all possibilities multiply the possibilities of all boxes and then subtract one to get the required number of possibilities.

Complete step-by-step solution:
To find the number of ways in which selection of one or more balls can be made from $10$ white, $9$ black and $7$ red balls is
We can either select none or “n” number of balls from a box, in this manner the selection ways for all boxes can be given as
Number of ways to select from set of $10$ white balls $ = 10 + 1 = 11$
Number of ways to select from set of $9$ black balls $ = 9 + 1 = 10$
Number of ways to select from set of $7$ red balls $ = 7 + 1 = 8$
Therefore total number of ways to select one or more balls
$
   = 11 \times 10 \times 8 - 1 \\
   = 880 - 1 \\
   = 879 \\
 $

Note: At last we have subtracted one from the total ways because that one stands for the selection of no balls from all three boxes, but as in the condition given in the question, we have to select one or more balls, not zero balls or none of the balls.