Palindromic Permutations of MISSISSIPPI

Palindromic permutations of MISSISSIPPI is an easy quant interview question on Combinatorics, reported to have been seen at Old mission.

Difficulty Easy Topic Combinatorics Reported at Old mission

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This combinatorics question focuses on counting symmetric permutations of a word with many repeated letters. It lives at the intersection of permutations of multisets and palindromic structure, a classic theme in quant prep and math interviews. Candidates must recognize the special constraints that symmetry imposes when letters are indistinguishable, and see how that shrinks the usual permutation space.

It trains your ability to handle counting with repeated elements while respecting an extra structural condition. You need to keep track of letter multiplicities, reason about feasibility before counting, and translate a verbal symmetry constraint into a precise combinatorial framework. This sharpens your discrete math intuition, pattern recognition, and careful bookkeeping under constraints.

This matters for quant interviews because many firms use similar puzzles to test precise reasoning under simple rules. It is a compact way to probe combinatorial thinking, error-checking, and comfort with structured counting, all essential in quantitative finance.

What it tests

Every palindromic arrangement of a multiset of letters is governed by the requirement that the string reads the same forwards and backwards. This means that for an odd-length palindrome, all letters except one must occur an even number of times, with the odd-count letter occupying the center. The arrangement of the first half of the palindrome completely determines the second half, due to the symmetry. The number of palindromic permutations is thus the number of ways to distribute half the occurrences of each letter (except possibly the central one) among the positions before the center, accounting for indistinguishable letters. This structure arises because mirroring enforces strict pairing, and the central position can only be filled by a letter with an odd count.

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