A bag contains 2 white b***s,3 black b***s and 4 red b***s. In how many ways can 3 b***s be drawn form the bag, if at least one black ball is to be included?
Bertie showed me how to get the permutations for this problem
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How many permutations of 3 b***s are there.
Well the first ball can be any of 3 colours, so can the second and so can the third so that is 3*3*3=27 BUT there are only 2 white b***s so one of these (that is WWW) does not exist
so 27-1=26 possible permutations.
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how many permutations contain no black one
Well the first one can be W or R,so can the 2nd and the third so 2*2*2 but once again they cannot all be white so we must take 1 away
so 8-1=7 permutations containing no black
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Therefore 26-7=19 permutations DO have at least one black ball. 
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Thanks Bertie. Some problems are so easy when some smart person shows you how to do them. 

