How Many Permutations Exist Where M Equals K on Cards?

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Discussion Overview

The discussion revolves around calculating the number of permutations of cards showing two values, M and K, where the number of M cards equals the number of K cards, given an even number of total cards, n. Participants explore the derivation of a formula for this scenario and discuss combinatorial concepts related to permutations and combinations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks to determine the number of permutations where the number of M cards equals the number of K cards, asking if a formula can be derived as a function of n.
  • Another participant suggests using the binomial coefficient to calculate the permutations.
  • A participant proposes the formula $$\binom n k = \frac{n!}{0.5n! \cdot 0.5n!}$$ for the case where k equals half of n, questioning its correctness.
  • Some participants affirm the correctness of the binomial coefficient approach and discuss the implications of indistinguishable objects in permutations.
  • There is a discussion about the need to account for overcounting in permutations when the order does not matter, leading to the use of division in the combinatorial calculations.
  • One participant explains that the number of ways to arrange k M's and k K's can be represented as choosing k positions from n, leading to the binomial coefficient $$\binom{n}{k}$$.
  • A suggestion is made that visualizing the problem through Pascal's Triangle could be helpful for understanding the combinatorial relationships involved.
  • Another participant elaborates on the derivation of the binomial coefficient, explaining the counting of arrangements and the need to account for indistinguishable items.

Areas of Agreement / Disagreement

Participants generally agree on the use of the binomial coefficient for calculating permutations in this context, but there are differing views on the treatment of indistinguishable items and the implications for counting methods. The discussion remains unresolved regarding the best approach to explain the derivation of the formula.

Contextual Notes

Some participants express uncertainty about the assumptions regarding distinguishability of the cards and the implications for counting permutations versus combinations. There is also mention of potential overcounting in arrangements, which complicates the derivation of a straightforward formula.

  • #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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