From a shuffled deck of 52 cards how many ways could you be dealt five cards containing 2 face cards, 2 aces and a five?
A bag contains 25 marbles of which 6 are blue. How many ways are there of choosing three marbles of which at least one must be blue?
What is the eighth term in the expansion of (x-(1/x))^16?
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Edited by kgj: 6/14/2013 3:34:46 AMFor the first one I don't have the deck of cards memorized so :( ____________________________________________________________ For the second one you have 25 marbles. 6 are blue. You're choosing 3 marbles. One has to be blue. So for the one blue versions: 6 blues X 19 Others X 18 Others [X 3 ways to arrange this, X2 again as you can flip the order of the Others] 6 blues X 5 Other Blues X 19 Others [X 3 ways to arrange this, X2 again as the order of the others can be flipped] 6 blues X 5 Blues X 4 Blues [Only one way to arrange] And add them all together. [This is assuming you count each blue as its own individual. If not then you just choose whatever many spaces you need from six for the blues, and whatever many spaces you need from 19 for the others] ___________________________________________________________- For this, you just apply the binomial expansion formula. I can't remember it off the top of my head.