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February 1, 2015

February 1, 2015

Posted by **CL** on Tuesday, April 13, 2010 at 8:43pm.

a) How many different bridge hands are possible?

b) How many different bridge hands have all pour aces in them?

c) How many different bridge hands have no aces in them?

I figured out part A, but am confused about the rest..

52nCr13 = 6.35 x 10^11 hands

- math - combinations -
**Reiny**, Tuesday, April 13, 2010 at 9:12pma) choose 13 from 52 = C(52,13) = 6.35 x 10^11

you had that

b) you have the four aces, that leaves 9 other cards from the remaining 48

so C(48,9) = 1677106640

c) imagine the deck with no aces, that would be 48 to choose from, but you still want 13 cards

C(48,13) = 1.929x10^11

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