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Calculus/Combination

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Expert: - 2/4/2011


Question
Hello,

Could you please help me with the following?
Q. Eight cards are selected with replacement, from a standard deck of 52 cards, with 12 picture cards, 20 odd cards and 20 even cards.
a. How many different sequences of eight cards are possible?(Ans: 5.346x10^13)
b. How many of the sequences in part a will contain three picture cards, three odd cards and two even-numbered cars. (Ans: 3.097x10^12)

Thank you
b.

Answer
Since there are 52 ways of secelecting each card, the answer to (a) is 52^8.
Note that 52^2 = 2,704.  Using this, 52^2^2 = 52^4 = 7,311,616.
That gives 379,456, so carry the 379.
Anyway, after doing all that, (7,311,616)² = 53,459,728,531,456, which is what is given.

For b, we have (12^3)(20^3)(20^2) = (1728)(8000)(400) = 5,529,600,000.
Now apparently order is important, and the number of orders are 8!/(3!3!2!),
which is 40,320/(6*6*2) = 560.  If we take 5,529,600,000/560, we get
3.09658E+12, which rounds to 3.097x10^12.

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