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  • In a deck of 52 playing cards, how many different 3-card hands can be formed such that at least 2 cards should be king and queen?  Option: 1</stro

In a deck of 52 playing cards, how many different 3-card hands can be formed such that at least 2 cards should be king and queen?

 

Option: 1

2,589


Option: 2

1,568


Option: 3

1,798


Option: 4

1,944


Answers (1)

best_answer

To find the number of different 3-card hands that can be formed from a deck of 52 playing cards such that at least 2 cards are king and queen, we can break down the problem into cases.

Case 1: 2 cards are king and queen, and the third card is any other card:

- Choose the king: 4 kings to choose from.

- Choose the queen: 4 queens to choose from.

- Choose the third card: 52 - 2 = 50 cards to choose from.

Therefore, the number of hands in this case is:4 \times 4 \times 50.
 

Case 2: All 3 cards are king and queen:

- Choose the suit for the 3 cards: 4 suits to choose from.
- Choose the 3 cards from the chosen suit: C(13, 3) ways to choose.
Therefore, the number of hands in this case is:4 \times C\left ( 13,3 \right ).
\mathrm{C(13,3)=13 ! /(3 ! \times(13-3) !)=13 ! /(3 ! \times 10 !)=(13 \times 12 \times 11) /(3 \times 2 \times 1)=286}.
 

Now we can substitute this value back into the expression:

\mathrm{4 \times 4 \times 50+4 \times C(13,3)=4 \times 4 \times 50+4 \times 286}.
Calculating this expression:
\mathrm{4 \times 286=1,144}.
Therefore, the expression \mathrm{4 \times C(13,3)} evaluates to 1,144 .

Finally, to get the total number of different 3-card hands, we sum up the number of hands from both cases:
Total number of hands \mathrm{=4 \times 4 \times 50+4 \times C(13,3)}.

\mathrm{C(13,3)=13 ! /(3 ! \times(13-3) !)=13 ! /(3 ! \times 10 !)=(13 \times 12 \times 11) /(3 \times 2 \times 1)=286}.
Now we can substitute this value back into the expression:
\mathrm{4 \times 4 \times 50+4 \times C(13,3)=4 \times 4 \times 50+4 \times 286}.

Calculating this expression:
\mathrm{4 \times 4 \times 50+4 \times 286=800+1,144=1,944}.
Therefore, the expression \mathrm{4 \times 4 \times 50+4 \times C(13,3)} evaluates to 1,944.

 

Posted by

Devendra Khairwa

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