We are supposed to use the Inclusion-Exclusion principle to solve: "how many bridge hands contain ex

Winigefx

Winigefx

Answered question

2022-06-14

We are supposed to use the Inclusion-Exclusion principle to solve: "how many bridge hands contain exactly 3 clubs, or exactly 5 diamonds or exactly 3 aces?" I know you do something like: |3 clubs|+|3aces|+|5dimonds| - |3clubs AND 3 aces| - |3clubs AND 5 diamonds| -...... +|all 3 intersected| But I'm really struggling with find the cases where the aces may or may not be a club/diamond.

Answer & Explanation

Esteban Johnson

Esteban Johnson

Beginner2022-06-15Added 15 answers

exactly  3   = ( 13 3 ) ( 39 10 ) = 181 823 183 256 exactly  5   = ( 13 5 ) ( 39 8 ) = 79 181 063 676 exactly  3   A = ( 4 3 ) A ( 48 10 ) = 26 162 863 584

exactly  3    and  5   = ( 13 3 ) ( 13 5 ) ( 26 5 ) = 24 212 433 960 exactly  3    and  3   A = ( 12 2 ) ( 3 2 ) A ( 36 8 ) with A + ( 12 3 ) ( 3 3 ) A ( 36 7 ) without A = 7 828 036 920 exactly  5    and  3   A = ( 12 4 ) ( 3 2 ) A ( 36 6 ) with A + ( 12 5 ) ( 3 3 ) A ( 36 5 ) without A = 3 191 048 784

exactly  3    and  5    and  3   A = ( 12 2 ) ( 12 4 ) ( 2 1 ) A ( 24 4 ) with A  and A + ( 12 2 ) ( 12 5 ) ( 2 2 ) A ( 24 3 ) with A  without A + ( 12 3 ) ( 12 4 ) ( 2 2 ) A ( 24 3 ) without A  with A + ( 12 3 ) ( 12 5 ) ( 2 3 ) A ( 24 2 ) without A  or A = 1 020 514 968

Inclusion-Exclusion says we have
( 181 823 183 256 + 79 181 063 676 + 26 162 863 584 ) ( 24 212 433 960 + 7 828 036 920 + 3 191 048 784 ) + 1 020 514 968 = 252 956 105 820
bridge hands with exactly 3   or exactly 5   or exactly 3   A.
Since there are ( 52 13 ) = 635 013 559 600 bridge hands, the probability of the specified hands is approximately 39.834756596275 %.

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