Question

A standard deck of cards consists of four suits (clubs, diamonds, hearts, and spades), with each suit containing $13$ cards (ace, two through ten, jack, queen, and king) for a total of $52$ cards in all. How many $7$ card hands will consist of exactly $2$ hearts and $3$ clubs?

Answer 1

Find the number of $7$ card hands that will consist of exactly $2$ hearts and $3$ clubs.

There are $13$ cards of hearts and clubs each.

Total cards in a deck are $52$.

Number of $7$ card hands that will consist of exactly $2$ hearts and $3$ clubs is ${}^{13}\mathrm{C}_2\times {}^{13}\mathrm{C}_3\times {}^{47}\mathrm{C}_2$.

Exactly $2$ heart cards can be selected in ${}^{13}\mathrm{C}_2$ ways.

Exactly $3$ heart cards can be selected in ${}^{13}\mathrm{C}_3$ ways.

Remaining card hands are $7-5=2$.

Remaining $2$ cards can be selected out of remaining $47$ cards in ${}^{47}\mathrm{C}_2$ ways.

Apply combination formula ${}^{n}C_r=\frac{n!}{(n-r)!·r!}$

$$\begin{gathered} {}^{13}\mathrm{C}_2\times {}^{13}\mathrm{C}_3\times {}^{47}\mathrm{C}_2=\frac{13!}{(13-2)!\times 2!}\times \frac{13!}{(13-3)!\times 3!}\times \frac{47!}{(47-2)!\times 2!} \\ \begin{array}{rcl}& & \begin{array}{cc}=& \frac{13!}{11!\times 2!}\times \frac{13!}{10!\times 3!}\times \frac{47!}{45!\times 2!}\\ =& 78\times 286\times 1081\\ =& 24114948\end{array}\end{array} \end{gathered}$$

Hence, the number of $7$ card hands will consist of exactly $2$ hearts and $3$ clubs are $24,114,948$.