A drum contains 4 red balls, 3 blue balls and 2 white balls. Three balls are removed (no replacement) what is the probability that there will be one of each?	Answers
A) 22 B) 21 C) 24 D) 25 E) 28
4*3*2 / 9*8*7.   There is a 4/9 chance of taking a red ball first; if you do, ther'es a 3/8 chance of taking a blue ball next, and if you get that far, there's a 2/7 chance of getting a white ball third.   Or you might get white first, blue second and red third: 2/9, followed by 3/8 and then 4/7. No matter what order it might occur, the multiplication will always be (4*3*2)/(9*8*7).   Since 4*2= 
<gadget id="calculator">4*2</gadget>
<output>8</output>
8 and 3 is a factor of 9, you can cancel out the entire top row to leave 1 / (3*1*7) ie. 1 chance in 21.   answer 21
<result>B</result>
A drum contains 4 red balls, 3 blue balls and 2 white balls. Three balls are removed (no replacement) what is the probability that there will be one of each?