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

For each colour we may choose any number of balls from $$0$$ up to the available stock. The choices for the three colours are therefore independent.

Number of choices for white balls: $$0,1,2,\dots ,10$$ ⇒ altogether $$10+1=11$$ possibilities.

Number of choices for green balls: $$0,1,2,\dots ,9$$ ⇒ altogether $$9+1=10$$ possibilities.

Number of choices for black balls: $$0,1,2,\dots ,7$$ ⇒ altogether $$7+1=8$$ possibilities.

Using the rule of product (multiplication principle), the total number of ordered triples $$(w,g,b)$$ that can be formed is
$$11 \times 10 \times 8 = 880.$$ Each triple represents a selection containing $$w$$ white, $$g$$ green and $$b$$ black balls.

The triple $$(0,0,0)$$ corresponds to selecting no ball at all, but the question asks for “one or more” balls. Hence we must exclude this single case.

Required number of selections:
$$880 - 1 = 879.$$

Option D which is: $$879$$