Chapter 9, Problem 12RE

Solution

To calculate: The numbers of different 13-card bridge hands that include all four aces and exactly one king.

The resultant answer is $708,930,508$ .

Given information:

The given statement says to find out numbers of different 13-card bridge hands that include all four aces and exactly one king.

Formula used: The number of combinations of n objects taken r at a time denoted by $C_{n_r}$ and given by $C_{n_r}=\frac{n!}{r!\left(n-r\right)!}$ .

Calculation:

Total number of cards in a deck: 52

13 cards are to be selected which includes all four aces and exactly one king.

Now note that out of 13 cards 4 cards of aces have already been selected and the fifth card of king can be selected in 4 ways.

Then the remaining card in the deck will become: $52-4-4=44$

Now from these 44 cards, 8 cards are to be selected: $C_{{44}_8}$

Rewrite the expression using the formula $C_{n_r}=\frac{n!}{r!\left(n-r\right)!}$ ,

  $\begin{array}{c}{C}_{{44}_8}=\frac{44!}{8!\left(44-8\right)!}\\ =\frac{44!}{8!36!}\end{array}$

Now expand the factorial,

  $\frac{44\cdot 43\cdot 42\cdot 41\cdot 40\cdot 39\cdot 38\cdot 37\cdot 36!}{36!\left(8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\right)}=177,232,627$

The total number of ways will become:

  $\begin{array}{c}{C}_{4_4}\cdot C_{4_1}\cdot C_{{44}_8}=\left(1\right)\cdot \left(4\right)\cdot 177,232,627\\ =708,930,508\end{array}$

Therefore, the total number of ways of selecting 13-card bridge hands that include all four aces and exactly one king is $708,930,508$ .