core29 min studyLesson 225 in Mathematics

Partitions and Ferrers diagrams

Count unordered integer decompositions using multiplicities, conjugate diagrams, and coefficientwise finite products.

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What you’ll understand

  • Distinguish integer partitions from compositions and set partitions.
  • Use Ferrers conjugation to exchange row and column constraints.
  • Derive generating products from independent part multiplicities.

Distributing eight identical units among unnamed piles produces an integer partition. Listing the piles in a different order does not create a new object, so a canonical nonincreasing list records the partition without duplication. Drawing one left-justified row for each part turns arithmetic restrictions into geometric ones. The same diagram can be read by columns, yielding a reversible transformation that explains equal counts between seemingly different families. Multiplicity generating functions provide a second description of these objects.

Order, identity, and a canonical list

A partition of a nonnegative integer n is a nonincreasing finite list of positive integers summing to n. The empty list is the unique partition of zero. The list (4,2,2)(4,2,2) is an integer partition of eight; changing its order does not produce another partition. A composition keeps order, so (2,4,2)(2,4,2) is different as a composition. A set partition is different again: it divides distinct labels into blocks, and block membership contains information absent from a list of integer sizes.

For each permitted positive size j, specify a nonnegative multiplicity. This produces a unique integer partition after sorting the parts. Allowing arbitrary multiplicity gives a geometric factor; allowing a part at most once gives a two-term factor. Although the unrestricted product contains infinitely many factors, its coefficient at degree n depends only on sizes up to n. That finite dependence justifies the product as a formal series without invoking convergence at a numerical point.

∑n≥0p(n)xn=∏j≥111−xj,∑n≥0q(n)xn=∏j≥1(1+xj)\sum_{n\ge0}p(n)x^n=\prod_{j\ge1}\frac1{1-x^j},\qquad \sum_{n\ge0}q(n)x^n=\prod_{j\ge1}(1+x^j)
p counts unrestricted partitions and q counts partitions into distinct parts. Both products are defined coefficientwise because all part sizes are positive.↔ Scroll wide equations horizontally.

Conjugating a Ferrers diagram

Draw a row of dots or cells for each part, left aligned with the longest row first. The conjugate partition has one part for every column, equal to the height of that column. Transposing twice restores the original diagram, proving bijectivity. The number of rows becomes the largest part and the largest part becomes the number of rows. Total size is preserved because the same cells are counted in a different direction.

Consequently partitions of n with at most k parts are equinumerous with partitions of n whose largest part is at most k. Equality conditions also transpose: exactly k rows correspond to largest part exactly k. These statements do not say that the two constrained families contain the same lists. For instance a long single row becomes a tall column. The diagram supplies the inverse map, which is stronger evidence than matching a short table of counts.

λj′=∣{i:λi≥j}∣,(λ′)′=λ,∣λ′∣=∣λ∣\lambda'_j=|\{i:\lambda_i\ge j\}|,\qquad (\lambda')'=\lambda,\qquad |\lambda'|=|\lambda|
The jth conjugate part is the height of column j. Transposition exchanges row and column restrictions while preserving the number of cells.↔ Scroll wide equations horizontally.

Distinct parts, odd parts, and justified cancellation

Partitions into distinct parts have the same count as partitions into odd parts. One formal derivation replaces each distinct-part factor by a quotient of two geometric numerators. Every even-indexed factor cancels, leaving reciprocal factors only for odd sizes. To interpret an infinite cancellation safely, work to a fixed degree and ignore all higher-degree terms. Only finitely many factors matter there, so the cancellation is an ordinary finite algebraic calculation at every requested coefficient.

There is also a reversible arithmetic explanation: in a partition into odd parts, express the multiplicity of each odd size in binary. Each nonzero binary digit becomes one part of that odd size multiplied by a power of two. The resulting parts are distinct because every positive integer has a unique odd factor times a power of two. Splitting each distinct part back into its odd factor recovers the original multiplicities. This bijection changes the number of parts, so it proves a total-size identity rather than a fixed-length identity.

∏j≥1(1+xj)=∏j≥11−x2j1−xj=∏r≥011−x2r+1\prod_{j\ge1}(1+x^j)=\prod_{j\ge1}\frac{1-x^{2j}}{1-x^j}=\prod_{r\ge0}\frac1{1-x^{2r+1}}
The identity holds coefficientwise as a formal product. Analytic evaluation and infinite-product convergence would require additional justification.↔ Scroll wide equations horizontally.

A diagram inside a three-by-four box

Find the conjugate of λ=(4,2,2). Then count all partitions of eight with largest part at most three.

  1. The four column heights of λ are 3,3,1,1. Thus its conjugate is (3,3,1,1), still of total size eight; its largest part is three because λ has three rows.

  2. For the counting question write the multiplicities of parts one, two, and three as a,b,c. They satisfy a+2b+3c=8, with every variable nonnegative.

  3. If c=0, b has five choices from zero through four. If c=1, b has three choices; if c=2, b has two. Higher c is impossible. The total is 5+3+2=10.

[x8]1(1−x)(1−x2)(1−x3)=10[x^8]\frac1{(1-x)(1-x^2)(1-x^3)}=10
Each factor records the multiplicity of one allowed part size.↔ Scroll wide equations horizontally.
The result

The conjugate is (3,3,1,1), and the constrained count is ten. By conjugation, ten also counts partitions of eight into at most three parts.

Explore the model

Interactive lab

Interactive lab · probability

Choosing distinct allowed parts

Pause and Reset, then set p=0.5 and n=5. Regard the five positions as part sizes 1,2,3,4,5, each included or excluded. Predict the number of subsets with exactly two parts by multiplying the two-success mass by 32. Then list their size sums separately.

Choosing distinct allowed partsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThere are ten two-part subsets, but their total sizes differ. The lab groups by number of selected parts, not by their weighted sum; the partition generating function must track that sum separately.

Check your understanding

What does conjugation do to partitions of n with exactly four parts?

Try it yourself

Apply the odd-to-distinct bijection to the partition (3,3,3,1,1,1,1,1) of fourteen, and verify the inverse.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The multiplicities are three for part three and five for part one. Write each multiplicity in binary.

Compare with the solution

Three is 1+2, so the three copies of three become parts three and six. Five is 1+4, so the five ones become parts one and four. Sorting gives the distinct partition (6,4,3,1), whose sum is fourteen. Conversely six splits into two threes, three stays one three, four splits into four ones, and one stays one one. This restores exactly the original odd-part partition.

Take it with you

  • Integer partitions discard order but retain each positive part multiplicity.
  • Ferrers conjugation exchanges number of parts with largest part.
  • Infinite formal products are valid when every fixed coefficient depends on finitely many factors.

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