document.write( "Question 1201199: How many ways to create a five card poker hand with 2 aces and 2 kings? \n" ); document.write( "
Algebra.Com's Answer #835470 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Answer: 1584\r
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\n" ); document.write( "\n" ); document.write( "Explanation:\r
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\n" ); document.write( "\n" ); document.write( "Let's say the first card we want is neither an ace nor a king.
\n" ); document.write( "There are 4 aces and 4 kings.
\n" ); document.write( "That's 4+4 = 8 cards total.
\n" ); document.write( "There are 52-8 = 44 cards that are neither an ace nor a king.
\n" ); document.write( "That's the number of choices we have for the first slot.\r
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\n" ); document.write( "\n" ); document.write( "Put another way: For any given suit, there are 13-2 = 11 cards that are neither an ace nor a king.
\n" ); document.write( "Four suits total yields 4*11 = 44 cards that are neither an ace nor a king. \r
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\n" ); document.write( "\n" ); document.write( "Now to count the number of ways to pick the two kings.
\n" ); document.write( "There are n = 4 kings to pick from and r = 2 slots to fill.
\n" ); document.write( "Order does not matter with poker hands.
\n" ); document.write( "This means we go for the nCr combination formula.
\n" ); document.write( "n C r = (n!)/(r!(n-r)!)
\n" ); document.write( "4 C 2 = (4!)/(2!*(4-2)!)
\n" ); document.write( "4 C 2 = (4!)/(2!*2!)
\n" ); document.write( "4 C 2 = (4*3*2!)/(2!*2!)
\n" ); document.write( "4 C 2 = (4*3)/(2!)
\n" ); document.write( "4 C 2 = (4*3)/(2*1)
\n" ); document.write( "4 C 2 = (12)/(2)
\n" ); document.write( "4 C 2 = 6
\n" ); document.write( "There are 6 ways to select the two kings.
\n" ); document.write( "This value can be found in Pascal's Triangle.\r
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\n" ); document.write( "\n" ); document.write( "Put another way: We have 4*3 = 12 permutations if order mattered.
\n" ); document.write( "But order doesn't matter so we divide by 2 to get 12/2 = 6 ways to pick the two kings.\r
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\n" ); document.write( "\n" ); document.write( "Since this value is fairly small, we can list all the ways to pick the two kings.
  1. KC,KD
  2. KC,KH
  3. KC,KS
  4. KD,KH
  5. KD,KS
  6. KH,KS
where,
\n" ); document.write( "KC = king of clubs
\n" ); document.write( "KD = king of diamonds
\n" ); document.write( "KH = king of hearts
\n" ); document.write( "KS = king of spades\r
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\n" ); document.write( "\n" ); document.write( "Order doesn't matter so a group like KC,KD is the same as KD,KC.\r
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\n" ); document.write( "\n" ); document.write( "Using practically identical logic, you should find that there are 6 ways to pick the two aces.\r
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\n" ); document.write( "\n" ); document.write( "Recap:
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  • 44 ways to pick a card that isn't an ace, and isn't a king either
  • 6 ways to pick two kings
  • 6 ways to pick two aces
That will give 44*6*6 = 1584 different five card poker hands consisting of 2 aces and 2 kings.
\n" ); document.write( "Order does not matter with poker hands.
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