Permutations and Combinations 4 Question 7
7. In how many ways can a pack of 52 cards be
(i) divided equally among four players in order
(ii) divided into four groups of 13 cards each
(iii) divided in 4 sets, three of them having 17 cards each and the fourth just one card?
$(1979,3$ М)
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Answer:
Correct Answer: 7. ((i) $\frac{(52) !}{(13 !)^{4}}$ (ii) $\frac{(52) !}{4 !(13 !)^{4}}$ (iii) $\frac{(52) !}{3 !(17)^{3}}$)
Solution:
- (i) The number of ways in which 52 cards be divided equally among four players in order
$ ={ }^{52} C _{13} \times{ }^{39} C _{13} \times{ }^{26} C _{13} \times{ }^{13} C _{13}=\frac{(52) !}{(13 !)^{4}} $
(ii) The number of ways in which a pack of 52 cards can be divided equally into four groups of 13 cards each $=\frac{{ }^{52} C _{13} \times{ }^{39} C _{13} \times{ }^{26} C _{13} \times{ }^{13} C _{13}}{4 !}=\frac{(52) !}{4 !(13 !)^{4}}$
(iii) The number of ways in which a pack of 52 cards be divided into 4 sets, three of them having 17 cards each and the fourth just one card
$ =\frac{{ }^{52} C _{17} \times{ }^{35} C _{18} \times{ }^{18} C _{17} \times{ }^{1} C _1}{3 !}=\frac{(52) !}{3 !(17)^{3}} $
