Originally posted by: cool_rashi
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Calculate the number (not the probability) of bridge hands which contain the following:
(Recall that a Bridge hand consists of 13 cards taken from a standard 52 card deck of playing
cards.)
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a) 5 clubs and 5 hearts
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b) 4 clubs and 3 cards in each of the other suits
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c) 7 cards of one suit
Could not reply earlier as I had not visited the forum in many weeks.
1a) There are 13 clubs, 13 hearts
5 clubs can be drawn from 13 clubs in 13C5 ways
5 hearts can be drawn from 13 hearts in 13C5 ways
Remaining 13-5-5=3 cards can be drawn from remaining 52-13-13 = 26 cards in 26C3 ways
No. of bridge hands = 13C5 X 13C5 X 26C3
2b) 4 clubs out of 13 clubs can be drawb in 13C4 ways
Each suit is of 13 cards.
So 3 cards from each of the other suit can be drawn in
13C3 X 13C3 X 13C3 ways
No. of bridge hands = 13C4 x 13C3 X 13C3 X 13C3
3c) There are 4 suits
There are 4 suits,
1 suit can be selected in 4 ways,
7 cards from this suit can be drawn in 13C7 ways
Remaining 13-7=6 cards have to be drawn from remaining 52-13 = 39 cards
This can be done in 39c6 ways
No. of bridge hands = 4 X 13C7 X 39C6
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