For a practice problem for my next test we were given this: Many bridge players
ID: 3236677 • Letter: F
Question
For a practice problem for my next test we were given this: Many bridge players evaluate their hands by the following system of point counts: give yourself one point for a jack, two for a queen, three for a king, and four for an ace. Furthermore, give yourself three points if there is a suit in which you have zero cards, two points if you have one card, and one point if you have two cards. How many bridge hands have a point count of zero?
I know that based on the scoring for bridge we need a hand of 13 cards with at least 3 cards of each suit, but I am unsure how to account for that. I am expecting to use n choose k (dont know how to type that notation sorry), but I'm not entirely sure how to. The only thing I can think of is if you do (24 choose 1) which would be picking the one remaining card after putting 3 from each suit (or 12 cards) and ignoring the face cards and aces. But I am pretty sure this is not right because it doesnt ensure that each suit has exactly three cards, ect. Any help would be appreciated.
Explanation / Answer
Points for each hand is called Honor Count (HC). You need the all possible hands with HC = 0.
Eliminate J, K, Q and A for all suits (16 in total) that leaves with 36 cards. Logic is simple since any one of the cards in you hand means HC > 0.
You can achieve zero HC by picking 3 in 3 of the suits plus 4 in remaining suit. For example, {3C,3H,3D,4S}.
Of the remaining 36 cards, each suit will have exactly nine and 3 can be chose in C(9,3) ways.
Once 12 cards are drawn, 24 remain and choosing any one would yield a hand with zero point count.
Required Hands = C(9,3)*C(9,3)*C(9,3)*C(9,3)*C(24,1)
= 84^4*24 = 1,194,891,264
Note: C(n,k) is no of ways of choosing k items from n objects and C(n,k) = n!/k!(n-k)!
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