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.

Explanation

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 balls at next 4 places, (3rd to 6th place) and 3 red balls at next 3 places (7th to 9th place), i.e., `B_(1)B_(2)W_(1)W_(2)W_(3)W_(4)R_(1)R_(2)R_(3)`, which can be done in `2! xx 4! xx 3!` ways. And total number of ways of arranging all 2 + 4 + 3 = 9 balls is `9!` Therefore the required probability is
`(2! xx 4! xx 3!)/(9!) = (1)/(1260)`