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In how many ways can a pack of 52 cards be divided equally among four players in order?

(a) (52 !)4 

(b) 4 × (13 !) 

(c) \(\frac{52!}{(13!)^4}\)

(d) None of these

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(c) \(\frac{52!}{(13!)^4}\)

This means that first player gets 13 cards, then second player gets 13 cards, then third player gets 13 cards and then fourth gets 13 cards. 

∴ First player gets 13 cards in 52C13 ways 

Second player gets 13 cards in 39C13 ways 

Third player gets 13 cards in 26C13 ways 

Fourth player gets 13 cards in 13C13 ways 

∴ Total number of ways = 52C13 × 39C13 × 26C13 × 13C13 

= \(\frac{52!}{39!13!}\times\frac{39!}{26!\times13!}\times\frac{26!}{13!13!}\times1=\frac{52!}{(13!)^4}\) 

Alternatively, 52 cards can be divided equally among 4 players in \(\frac{52!}{(13!)^4\times4!}\) ways. 

But as here order is important, hence required number of ways = \(\frac{52!}{(13!)^4\times4!}\) = \(\frac{52!}{(13!)^4}\)

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