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A standard deck of playing cards consists of 52 cards. Each card has a rank and a suit. There are 13 possible ranks (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A), 4 possible suits (spades, clubs, hearts, diamonds), and 13 cards for each suit (one for each rank). How many different hands of 7 cards…

a. contain 4 cards of one rank and 3 cards of a second rank?

b. contain 3 pairs of 3 different ranks and a single card of a fourth rank?

My attempt

a)$P(13,2)*C(4,4)*C(4,3)$ ways

b) $C(13,3),C(4,2),C(4,2),C(40,1)$

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  • $\begingroup$ Your first answer is correct. You should have multiplication signs in your second answer. Also, you are missing a factor of $\binom{4}{2}$. $\endgroup$ – N. F. Taussig Apr 26 '18 at 3:01
  • $\begingroup$ But I already have $(4,2)$ or can you please explain $\endgroup$ – tien lee Apr 26 '18 at 3:05
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    $\begingroup$ You need to select a pair for each of the three ranks. $\endgroup$ – Graham Kemp Apr 26 '18 at 3:19
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You do not distinguish hand by permutation of cards.

Firstly, you wish to count ways to: select four from four suits for one from thirteen rank, and three from four suits for one from the twelve remainining ranks.

$$\binom 44\binom {13}1~\binom 43\binom {12}1$$

That is, there are $\mathrm C(4,4)C(13,1)C(4,3)C(12,1)$ ways to select seven cards that contain 4 cards of one rank and 3 cards of a second rank .


Secondly, you wish to count ways to select two from four suits each for three from thirteen ranks, and one from four suits for one from the ten remaining ranks. $$...$$

That is, there are so many ways to select seven cards that contain 3 pairs of 3 different ranks and a single card of a fourth rank.

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a) Your method is correct. Interpretation: select (order is important) $2$ ranks from $13$ available (later the first rank will have $4$ cards and the second rank will have $3$ cards). Afterwards (now you are selecting from the two ranks) select (order is not important) $4$ cards from $4$ available in the first rank and select (order is not important) $3$ cards from $4$ available in the second rank. Hence: $$P(13,2)C(4,4)C(4,3).$$

Alternatively, select (order is not important) $1$ rank from $13$ available, then for this rank select (order is not important) $4$ cards from $4$ available. Then select (order is not important) $1$ rank from the rest $12$ left, then for this rank select (order is not important) $3$ cards from $4$ available. Hence: $$C(13,1)C(4,4)C(12,1)C(4,3).$$

b) Your method is almost correct. Interpretation: select (order is not important) $3$ ranks from $13$ available (later each will have a pair). Afterwards (now you are selecting for the three ranks) select (order is not important) $2$ cards from $4$ available in the first rank, then select (order is not important) $2$ cards from $4$ available in the second rank and select (order is not important) $2$ cards from $4$ available in the third rank. Afterwards select (order is not important) $1$ rank from the rest $10$ left. Afterwards select (order is not important) $1$ card from $4$ available for this rank. Hence: $$C(13,3)C(4,2)C(4,2)C(4,2)C(10,1)C(4,1).$$

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