Since this problem was discussed a lot in the recent 10-Questions Quiz, I’ve tried to put forward a conceptual brute-force approach by @gregmat, and a generalization explained by @kautsgre.

# The brute force approach

Let’s see how many permutations we get.

Since we have 4 distinct dishes, the number of permutations is 4! = 24.

Column A | Column B | Column B | Column D |
---|---|---|---|

ABCD | BACD | CABD | DABC |

ABDC | BADC | CADB | DACB |

ACBD | BCAD | CBAD | DBAC |

ACDB | BCDA | CBDA | DBCA |

ADBC | BDAC | CDAB | DCAB |

ADCB | BDCA | CDBA | DCBA |

Out of these, which permutations get at least one dish at the correct place?

Column A | Column B | Column C | Column D |
---|---|---|---|

ABCD |
BACD |
CABD |
DABC |

ABDC |
BADC | CADB | DACB |

ACBD |
BCAD |
CBAD |
DBAC |

ACDB |
BCDA | CBDA |
DBCA |

ADBC |
BDAC | CDAB | DCAB |

ADCB |
BDCA |
CDBA | DCBA |

What do you notice about the non-highlighted permutations?

They serve all dishes at the wrong places, exactly what we need to count to get the probability!

- Here, we use a concept called derangements.
- Derangement means a permutation in which no element is in its original/fixed place.
- In our assumption it means- A cannot be at position 1, B cannot be at position 2, C cannot be at position 3 and D cannot be at position 4.
- Clearly, we get 9 derangements.

The required probability is

# Generalization if you need to work with larger numbers

Here, we use the derangement formula, described by @kautsgre on Combinatorics question Using Sets

Breaking it down step by step:

Combine the terms:

Simplify:

Cancel out the common factor:

The required probability is

Thanks for reading!