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| Question 1201482:  This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.
 How many hands contain 2 kings?
 
 Answer by math_tutor2020(3817)
      (Show Source): 
You can put this solution on YOUR website! Answer: 103,776
 This number is slightly smaller than 104 thousand.
 
 Explanation:
 
 We can use the nCr combination formula
 Therefore we have x*y = 6*17296 = 103,776 different five-card hands that consist of exactly 2 kings. Order doesn't matter.x = 4C2 = 6 ways to select the two kings in any order (see the next section below for the steps)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)
 
 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.
 https://mathworld.wolfram.com/Combination.html
 https://www.mathsisfun.com/combinatorics/combinations-permutations.html
 https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php
 
 Here is another question about card hands
 https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1201443.html
 
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