Question 1194109: 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 3 kings and 3 queens?
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The card deck has 4 kings and 4 queens.
There are 4 ways to pick a king and toss it aside, which means you keep the other 3 kings. The same applies to the queens as well.
So far we have 4*4 = 16 ways to pick the three kings and three queens. The order doesn't matter.
The other 52-4-4 = 52-8 = 44 cards aren't a king nor a queen. This represents the number of ways to pick that seventh card.
We have 16*44 = 704 ways to form a seven-card hand that has exactly three kings and three queens.
Answer: 704
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The requirement is to choose 3 of the 4 kings, AND 3 of the 4 queens, AND 1 of the other 44 cards. Since all three requirements must be met, multiply the numbers of ways of making each of those selections:
ANSWER: 704
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