I have regular deck of playing cards and I want to find out how many color combinations there are in the deck. The cards fall into two categories: red and black. Here are the givens. Total $52$ items; $2$ distinct, repetition; $26$ black; $26$ red; Cards like $7$ of diamonds and king of hearts are red, while cards like the ace of spades and jack of clubs are black.

  • $\begingroup$ You won't get any answers unless you show what you've already did. This isn't somewhere to go post your homework problems. $\endgroup$ – Andrew Tawfeek Sep 26 '17 at 1:36
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    $\begingroup$ This isn't a homework problem. I'm just curious of what it would be and if there is a formula of it. $\endgroup$ – TH3 F145H Sep 26 '17 at 1:42
  • $\begingroup$ Are you asking how many sequences of red and black cards there are or how many times the pattern red-black-black-red-red-black can appear in the sequence? $\endgroup$ – N. F. Taussig Sep 26 '17 at 8:39


Just focus on one color (say, red), and think about the position of the 26 red cards in the deck (positions range from 1 to 52). That is, the red cards take up 26 of the 52 possible positions ... so how many sets of 26 numbers out of 1 through 52 are there?

  • $\begingroup$ I'm a little confused in what you mean by how many sets of 26 numbers out of 1-52 are there. $\endgroup$ – TH3 F145H Sep 26 '17 at 1:44
  • $\begingroup$ OK, suppose the first 26 cards are all red. Then they take up positions 1 through 26. So this would correspond to the set $\{ 1,2,3,..., 26 \}$. Now suppose the first 25 cards are all red, and also the very last one. That would correspond to the set $\{ 1,2,3,..., 25, 52 \}$. And if every second card is red, we get the set $\{ 2,4,6,...,48,50,52 \}$. In other words, every color combination corresponds to a selection of 26 numbers out of the numbers 1 through 52. $\endgroup$ – Bram28 Sep 26 '17 at 1:49
  • $\begingroup$ I understand. This would be a very big number. Impossible to count. What method can we use to get this number? Factorial? If so, I know we can't just use 26! I'm stuck here not knowing what to do. $\endgroup$ – TH3 F145H Sep 26 '17 at 2:01
  • $\begingroup$ @TH3F145H Yes, a very big number indeed! OK, so what I had hoped my HINT would point you towards is the 'choose' function ... if you are unfamiliar with that function, look it up. $\endgroup$ – Bram28 Sep 26 '17 at 2:05
  • $\begingroup$ Is it C (52, 26)? $\endgroup$ – TH3 F145H Sep 26 '17 at 2:14

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