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Setup

  • We have 4 unique cards
  • I am trying to find the probability that none of them are in the same position after shuffling

The Given Solution

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My Questions

  • They reason that there are 4 permutations with at least three cards in position, to me this doesn't make sense if there are 3 cards in position, then the last one must also be in position, therefore there is only one way to have all four in position
  • In the final calculation, I don't understand their method of subtraction, they have an overcounted number 4!, and are trying to remove all the permutations that have been counted for, though I don't really understand the way they grouped the numbers with parenthesis and why they did that.

Hopefully someone can help me understand the above questions. Thank you!

cuppajoeman
  • 319

1 Answers1

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I advice you to have a look at the so-called principle of inclusion/exclusion.

In this answer I will not justify the steps in the proof above, but will only apply this principle.

If you want a more profound knowledge on the principle then have a look at this answer.

Number the cards with $1,2,3,4$ and for $i=1,2,3,4$ let $A_i$ denote the number of orders such that card $i$ is on the same position after shuffling.

Then to be found is the cardinality of $A_1^{\complement}\cap A_2^{\complement}\cap A_3^{\complement}\cap A_4^{\complement}=(A_1\cup A_2\cup A_3\cup A_4)^\complement$.

The total number of orders equals $4!$ so that:$$|(A_1\cup A_2\cup A_3\cup A_4)^\complement|=4!-|A_1\cup A_2\cup A_3\cup A_4|$$

With the techniques of inclusion/exclusion and symmetry we find that this equals:$$4!-\binom41|A_1|+\binom42|A_1\cap A_2|-\binom43|A_1\cap A_2\cap A_3|+\binom44|A_1\cap A_2\cap A_3\cap A_4|=$$$$4!-4\cdot3!+6\cdot2!-4\cdot1!+1\cdot0!=9$$

drhab
  • 151,093