How Many Permutations Exist Where M Equals K on Cards?

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SUMMARY

The discussion centers on calculating the number of permutations of n cards, where each card can display either M or K, with an equal number of each (n/2 M's and n/2 K's) when n is even. The formula derived for this scenario is the binomial coefficient, expressed as 𝔋{n}{k} = \frac{n!}{(n/2)! \cdot (n/2)!}. This formula accounts for the indistinguishability of M's and K's by dividing the total permutations by the factorial of the counts of each type. The discussion also emphasizes the importance of distinguishing between permutations and combinations in combinatorial problems.

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  • #31
JohnnyGui said:
Hello,

I have been trying to solve this problem but I can't seem to find a way.

Given are ##n## cards and each card can show one of two values: M or K.

How many possible permutations are there in which there are as many cards with M as there are with K? Given that ##n## is an even amount of cards.

Is it possible to derive a formula for this as a function of ##n##? How does one deduce this?

This way makes sense to me ( a rephrasing of omeone else's answer, I think Dr Claude's ) : Assume you need to go from point A to point B along a grid system , where you must go ,say, north (M) j times and east(K) j times in order to arrive at B, i.e. n=2j. How many ways can you do this trip? Once the j places where you make a turn east(north) fully determine the rest of the trip.
This gives you a way of counting paths where M=K. The total number of paths is straightforward.
 
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  • #32
StoneTemplePython said:
This is pretty far astray from the original post that this thread is under. I.e. your original question was asked and answered. Some follow-ups, also asked and answered. Now you have a question about inference -- this requires a new thread, at a minimum. Your line of thinking here doesn't make sense to me. With a large enough number of tosses we should be able to estimate probability of heads up to any amount of (in the real world, reasonable) precision. There are a lot of different approaches, and ways to frame the inference problem.

Personally, I think you need to study probability theory first, then revisit these questions in a few months.

I was talking about when those large number of tosses are divided into small number of tosses, each being a trial, and how one can interpret these small trials to deduce the individual chance.
It was my intention to ask the question in my OP as a base that leads to this question. Making a new thread for every question that I have regarding this would seem ineffective to me. I got it eventually figured out though, so nevermind.

WWGD said:
This way makes sense to me ( a rephrasing of omeone else's answer, I think Dr Claude's ) : Assume you need to go from point A to point B along a grid system , where you must go ,say, north (M) j times and east(K) j times in order to arrive at B, i.e. n=2j. How many ways can you do this trip? Once the j places where you make a turn east(north) fully determine the rest of the trip.
This gives you a way of counting paths where M=K. The total number of paths is straightforward.

This is a creative way to think about it. It helped me deduce the same formula again. Thanks!
 

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