In how many ways can you choose $5$ cards from a deck of $52$ playing cards such that exactly $2$ denominations are same and all suits are available?

My attempt : $^{13}C_4 \times \text{} ^4 C_1\times\text{} 4!\times3 $

Choosing four cards. Choose Card to be repeated. Give 4 suits to your 4 denominations. Give remaining card one of the other suits.

But the correct answer given is : $^{13}C_5\times\text{}^4C_1\times108$

Where am I wrong?

  • $\begingroup$ I'm not sure I understand what "all suits are available" means. I would go 13 ways to choose the repeated denomination, times 6 ways to choose the 2 cards from that denomination, times 48 ways to choose the third card, times 44 ways to choose the 4th, times 40 ways to choose the 5th, divided by 6 permutations of the last 3 cards. $13\times48\times44\times40$. $\endgroup$ – Gerry Myerson Apr 25 '14 at 12:22

In your counting you miss the cases that the repeated suit belongs to two of the non-repeating denominations. For example you miss the case (denote by A,B,C,D the suits):

$$1A, 2A, 3B, 4C, 4D$$

  • $\begingroup$ Bless You.${}{}$ $\endgroup$ – evil999man Apr 25 '14 at 12:55

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