Balls are drawn one-by-one without replacement from a box containing 2 black 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black. 4 white and 3 red.

To draw 2 black, 4 white and 3 red balls in order is same as arranging two black balls at first 2 places, 4 white at next 4 place, (3^rd to 6^th place) and 3 red at still next 3 places (7^th to 9^th place), I.e., B₁ B₂ W₁ W₂ W₃ W₄ R₁ R₂ R₃, which can be done in 2! x 4! x 3! Ways. And total ways of arranging all 2 + 4 + 3 = 9 balls is 9!

∴ Required probability = 2! x 4! x 3!/9! = 1/1260