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I have two sets (set A, set B) of three letters (X, Y, D). Each letter can have one of two colors (red or blue). How many different combinations are possible. I would like the equation to determine how many.

You could view the problem as follows: - SET A: [X, Y, D] - SET B: [X, Y, D] Each letter can be either red or blue. One possible combination is - SET A: [red X, red Y, blue D] - SET B: [red X, blue Y, red D]. I would like an equation to determine the total of all the different possible combinations.

Just for clarity, another possible combination is:

  • SET A: [red X, red Y, red D]
  • SET B: [red X, blue Y, blue D].
Daniel Parsons
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    Apply multiplication principle to the following set of steps: Choose if setAitemX is red or blue (two options). Choose if setAitemY is red or blue (two options)... ... choose if setBitemD is red or blue (two options). Multiply the number of options available at each step to get the total number of arrangements possible. See rule of product on wikipedia. – JMoravitz Jul 30 '18 at 16:50
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    The fact that there are "two sets" and within each set the letters present match is superfluous information. All that matters is the number of items which we are coloring, in this case six items total, and the number of available colors, in this case two colors. – JMoravitz Jul 30 '18 at 16:52
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    $2$ options for choosing the color of setAitemX, $2$ options for choosing setAitemY, etc... multiplying the number of choices gives $2\times 2\times 2\times\cdots \times 2 = 2^6$ is $64$ total possibilities. The whole point of elementary combinatorics is to learn how to count without brute force, never answer questions laboriously by listing out every possible combination if you can help it. – JMoravitz Jul 30 '18 at 17:13
  • Thank you very much for your solution. I believe it is what I am looking for however I still need to list everything out laboriously as I have a sceptical mind by nature. Probably a curse but that is how it is with me. I want to see each of the 64 possibilities listed first before I can trust the mathematical equation. I know I should simply trust the undeniable purity of mathematics but I still feel uncomfortable without having solid evidence that the equation works. Like seeing one orange being added to a basket that already contains an orange, which then gives the result of two oranges. – Daniel Parsons Jul 30 '18 at 17:21
  • Read again the page about the rule of product on wikipedia. Search around this site for more information on it, additional examples, and probably somewhere a proof. A sketch of a proof: consider only two steps at a time, making a table, listing the options for the first step horizontally along the top and the options for the second step vertically along the side. You have then a rectangular grid, showing each possible way the two steps could be completed simultaneously. – JMoravitz Jul 30 '18 at 17:26
  • Go on to include a third step or further by taking the combined outcomes of the earlier steps as your new horizontal and the newly added step as the vertical, again a rectangle which you can count how many total spots in the grid there are using multiplication of the length of the grid by the height of the grid. This, along with induction will give the familiar formula $|A\times B\times \dots \times N|=|A|\times |B|\times \dots \times |N|$ where the $\times$ on the left is the cartesian product while the $\times$ on the right is the usual multiplication of numbers. – JMoravitz Jul 30 '18 at 17:28
  • I have taken your advice and I am following up on it. If you are looking for a more challenging problem then how about finding all the permutations if the order of the letters can move in each set. For example: Set A: [red Y, red D, blue X] and set B: [red D, blue X, blue Y]. In the meantime I will take a deeper look into the rule of product. I am considering two steps at a time and mapping out all the possible ways. Thank you again for pointing me in the right direction. – Daniel Parsons Jul 31 '18 at 00:19
  • That would be just repeated applications of the rule of product. Choose how each element is colored (as discussed earlier, can be done in $2^6$ ways). Then choose in what order the elements appear in the first set (can be done in $3!=3\cdot 2\cdot 1 = 6$ ways, also seen by rule of product). Then choose in what order the elements appear in the second set (also 6 ways), giving $2^6\cdot 6\cdot 6$ total ways. – JMoravitz Jul 31 '18 at 00:22
  • After studying Elementary Combinatorics I understand it now as follows. Each Set has 3 letters that are either blue or red. In my original question there are two sets. (Set A, Set B). For any given set there are 8 possible outcomes. [red X, red Y, red D], [blue X, blue Y, blue D], [red X, red Y, blue D], [red X, blue Y, blue D], [blue X, blue Y, red D], [blue X, red Y, red D], [red X, blue Y, red D], [blue X, red Y, blue D]. – Daniel Parsons Jul 31 '18 at 01:43
  • There are 8 possibilities and two sets. Set A = m, Set B = n. As each set has 8 possibilities, 'm = 8' and 'n = 8'. In Combinatorics (m * n = product) so (8 * 8 = 64). Pardon my ignorance but prior to posting my question I had never even heard of Combinatorics. OK, I am now at peace with the universe. – Daniel Parsons Jul 31 '18 at 01:44
  • Enjoy your journey through one of my favorite subjects. The early stuff is quite fun and has several applications, particularly in probability (e.g. calculating odds of poker hands). Be warned that some questions, although easy to ask, can be very difficult to answer. A famously difficult problem is counting the number of ways one can make a tower of six identical $2\times 4$ lego bricks (or more generally, a tower of $n$ $2\times 4$ lego bricks), something which LEGO themselves got wrong the first time they attempted the problem. – JMoravitz Jul 31 '18 at 01:49
  • Here is an elementary Combinatorics, counting, and The Multiplication Principle webpage that explains the entire equation. http://dp-portfolio.com/tutorials/combinatorics – Daniel Parsons Aug 01 '18 at 14:10

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