A box contains 5 white balls, 3 red balls and 2 green balls.

The balls are taken from box. With replacement and taking a total of 3 balls,

P(all balls have different colour)=(5/10)x(3/10)x(2/10) x 3! =0.18

This answer is the same as at the back of the book. My explanation is that a white, red or green ball must be taken and there are 3! ways of arranging the 3 scenarios.

Is this correct????

Then secondly,

Without replacement and taking a total of 5 balls

My wrong method:

P(exactly 2 balls are white) = (5/10)x(4/9)x(5/8)x(4/7)x(3/6) x 5!

This probability cannot be correct since it is above 1. What is the loophole that I did not see???

I dont need the answer to this question (I can solve using counting total number of cases) but I want to know the loophole in my thinking.

Thanks,