Question 1201482
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Answer: <font color=red size=4>103,776</font>
This number is slightly smaller than 104 thousand.


Explanation:


We can use the nCr combination formula<ul><li>x = 4C2 = 6 ways to select the two kings in any order (see the next section below for the steps)</li><li>y = 48C3 = 17296 ways to select the three other cards that aren't a king in any order (eg: Queen of diamonds, Jack of clubs, 9 of hearts)</li></ul>Therefore we have x*y = 6*17296 = <font color=red size=4>103,776</font> different five-card hands that consist of exactly 2 kings. Order doesn't matter.


Example hand: 
King of clubs, King of diamonds, 2 of spades, 3 of hearts, 8 of clubs
The order does not matter.


Since order doesn't matter, we use the nCr formula (not the nPr formula) to compute each part. 
I'll show the steps for each portion.
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Selecting the 2 kings
n = 4 kings total
r = 2 selections
n C r = (n!)/(r!(n-r)!)
4 C 2 = (4!)/(2!*(4-2)!)
4 C 2 = (4!)/(2!*2!)
4 C 2 = (4*3*2!)/(2!*2!)
4 C 2 = (4*3)/(2!)
4 C 2 = (4*3)/(2*1)
4 C 2 = 12/2
4 C 2 = 6
There are 6 ways to pick the two kings in any order.
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Selecting the 3 other cards that aren't a king
n = 52-4 = 48 cards that aren't a king
r = 3 slots to fill
n C r = (n!)/(r!(n-r)!)
48 C 3 = (48!)/(3!*(48-3)!)
48 C 3 = (48!)/(3!*45!)
48 C 3 = (48*47*46*45!)/(3!*45!)
48 C 3 = (48*47*46)/(3!)
48 C 3 = (48*47*46)/(3*2*1)
48 C 3 = 103776/6
48 C 3 = 17296
There are 17296 ways to pick the three other cards that aren't a king in any order.
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Further reading about the nCr formula.
The third link is a calculator that specializes in this formula.
<a href = "https://mathworld.wolfram.com/Combination.html">https://mathworld.wolfram.com/Combination.html</a>
<a href = "https://www.mathsisfun.com/combinatorics/combinations-permutations.html">https://www.mathsisfun.com/combinatorics/combinations-permutations.html</a>
<a href = "https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php">https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php</a>


Here is another question about card hands
<a href = "https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1201443.html">https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1201443.html</a>
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