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If eight cards are selected WITH replacement from a standard pack of 52 cards, where there are 12 picture cards, 20 odd-numbered cards and 20 even-numbered cards. How many sequences will contain 3 picture cards, 3 off-numbered cards and 2 even-numbered cards?
I have figured out that there are 52^8 = 5.346x10^13 different sequences. I just can't figure out how to get the answer for the question.
Would it not be 12^3x20^3x20^2 = 5.5296x10^9?
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Ah I see, thank you.
I think that formula counts sequences with repeated cards more than once.
Correcting it would be messy though, since a compensating term would have a similar problem, in that it would overcount sequences with more than one repeat.
Why did the vector cross the road?
It wanted to be normal.
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