Chapter 9.6, Problem 65E

a.

To determine

Find the ways can you get a full house.

Answer to Problem 65E

  $3744$

Explanation of Solution

Calculation:

For a full house three cards are of same kind(kind refers to whether it is an ace, king, queens etc.. ) and other two are same but different from the three cards.

First we need to choose a kind from $13$ kinds for the first $3$ cards that can be done in ${}_{13}{C}_3$ ways. Then we need to choose other kind from the remaining $12$ for the other $2$ cards. This can be done in ${}_{12}{C}_2$ ways.

After selecting the kind we need to choose the suit of the cards. For the $3$ cards that are same we need to choose $3$ suits out $4$ of suit, which can be done in ${}_4{C}_3$ . for the remaining same cards we need to choose suits from the $4$ available suits, which can be done in ${}_4{C}_2$ ways.

By the fundamental counting principle, total number of full houses are

  ${}_{13}{C}_1{\times }_{12}{C}_1{\times }_4{C}_3{\times }_4{C}_2=\frac{13!}{12!1!}\times \frac{12!}{11!1!}\times \frac{4!}{1!3!}\times \frac{4!}{2!2!}$

  $\begin{array}{l}=13\times 12\times 4\times 6\\ =3744\end{array}$

Hence, the total number of full houses are $3744$ .

b.

To determine

Find the ways can you get five-card combination containing two jacks and three aces.

Answer to Problem 65E

  $24$

Explanation of Solution

Calculation:

Two jacks can be selected from jacks in a pack of cards in ways.

3 aces can be selected from aces in a pack of cards in ways.

By the fundamental counting principle, total number of combinations are

  $\begin{array}{l}{}_4{C}_3{\times }_4{C}_2=\frac{4!}{1!3!}\times \frac{4!}{2!2!}\\ =4\times 6\\ =24\end{array}$

Hence, the total number of combinations are $24$ .