Respuesta :
Answer:
123,552 ways
Step-by-step explanation:
Solution:-
- A deck of cards has a total of 52 cards. There are 4 suits/ranks in total: Hearts, Spades, Diamonds, and Clubs. There are 13 cards in each suit/rank. So, 4 suits/ranks with 13 cards in each = 4 x 13 = 52 cards ( complete deck ).
- Each suit/rank is divided into similar category of cards. These categories are:
Numbered cards ( From "A or 1 " to "10" ) = 10 cards
Jack = 1 card
Queen = 1 card
King = 1 card
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Total = 13 ordinals
- We are given 5 cards. The 5 cards in such a manner:
2 cards from one rank/suit
- Since any particular rank or suit ( Hearts, Diamonds, Clubs, or Spades) has 13 cards. We are first going to "select" 2 cards from 13 ordinals of a rank. We will use combinations " C " to determine the number of ways:
[tex]13 C 2 = \frac{13!}{2!*11!} = 78 ways[/tex]
Now we need to do is pick 2 cards from each suit for each ordinal ( Total of 4 suits ) for the first pair. We can determine it:
[tex]4 C 2 = \frac{4!}{2!*2!} = 6 ways[/tex]
Similarly, for the second pair we pick 2 cards from each suit ( Total of 4 suits ) for each ordinal.
[tex]4 C 2 = \frac{4!}{2!*2!} = 6 ways[/tex]
- We are done with 2 cards for 2 different ordinals. To select the last card in the 5 card-hand. We are left with 11 ordinals in each suit/rank. From which we choose 1 card.
[tex]11 C 1 = \frac{11!}{1!*10!} = 11 ways[/tex]
- And from 4 ordinals we need one rank/suit:
[tex]4 C 1 = \frac{4!}{1!*3!} = 4 ways[/tex]
- All the events listed above are dependent events. So the total number of ways to hand out the 5 cards such that there are 2 pairs belonging from 2 ranks and last card from 3rd rank. We have:
= 13C2 * 4C2 * 4C2 * 11C1 * 4C1
= 78 * 6 * 6 * 11 * 4
= 123,552 ways
Answer:
First we decide what ranks we will use. Then we will pick suits for all of the cards.
We choose the two paired ranks in ${13\choose 2}=78$ ways and the remaining rank in ${11\choose 1}=11$ ways. Then we choose the suits for these cards in ${4\choose 2}{4\choose 2}{4\choose 1}=144$ ways. This gives a total of $78\cdot 11\cdot 144 = \boxed{123552}$ two pair hands.
