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- 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)
  Therefore we have x*y = 6*17296 = 103,776 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. 
----------------- 
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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