Specific combinatorial cases?

Question

Okay, so I was thinking last night, and if you think about two circles.

And, if we imagine that these circles are specific. The one on the left is circle one and the one on the right is circle two. If we have two colours that could fill these circles, then there are four combinations of colours. What we find is that 2C2 does not work.

Or to make it less abstract, imagine there are two booths, specific booths, where you can’t see the booth operators face. The booth operator in booth one could either be male or female, and the booth operator in booth two could either be male or female. You must go to each booth to finish your business. Well, there are four combinations of possibilities.

Now what I’ve found is that to compute the combinations of any specific combinatorial cases, we take x^n where x is the number of specific cases, and n is the number of variations of each case. The probability of the case being of a variation is 1/n. So, in the booth operator case, it is 2^2, and the chances are 50% for each variation.

Could someone chime in on this please?

Is this how it works for specific combinatorial cases like this? I've thought up to 5 circles and 2 colours, and the combinations are equal to 32.

Edit: Sorry, I should explain the probibility thing. The probability of booth operator one being female is 50%, but the probability of each specific combination is 1/x^n

Ross Millikan · Answer 1 · 2017-03-29T15:03:29.313

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You are talking about the rule of product or multiplication principle. If you have $a$ choices for one thing and $b$ choices for another, the number of choices for the pair, assuming no other restrictions, is $ab$. This extends to any finite number of things. It gets used very often in combinatoric questions here.

edited Mar 29 '17 at 15:03

answered Mar 29 '17 at 14:55

Ross Millikan

374,822

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The point of the principle is that you can make each choice independently. For each of the $a$ choices for the first you get $b$ choices for the second. Add up $a\ b$s and you get $ab$ – Ross Millikan Mar 29 '17 at 15:04
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It comes up in the proof that there are $n!$ ways to put $n$ things in order. You have $n$ choices for the first one, $n-1$ choices for the second, $n-2$ for the third, and so on. A question where this comes up several times is this one – Ross Millikan Mar 29 '17 at 15:13
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No, for three circles, each of which may have two colors, you get $2 \cdot 2 \cdot 2=2^3=8$ possibilities. As you said in the original post, if you have five circles you get $2^5=32$ – Ross Millikan Mar 29 '17 at 15:15
This comes up several times in There are apparently 30723072 ways to draw this flower. But why? – Ross Millikan Mar 29 '17 at 15:20

Specific combinatorial cases?

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