Questions: Integer Partitions and Partition Functions

Question 1

How many partitions does the integer 5 have?

• A. 5
• B. 6
• C. 7
• D. 8

Answer: (C)

The partitions of 5 are: 5; 4+1; 3+2; 3+1+1; 2+2+1; 2+1+1+1; 1+1+1+1+1 — seven in total, so p(5) = 7. A common error is to count only the 'most obvious' decompositions (into two parts, etc.) and miss the cases with many small parts or the partition consisting of the number itself. Listing systematically by largest part avoids omissions.

Question 2

The Ferrers diagram of the partition 4+2+1 of 7 is conjugated (reflected along its main diagonal). Which partition results?

• A. 1+2+4 — the parts of the original, reversed in order
• B. 3+2+1+1 — reading column heights left to right from the original diagram
• C. 7 — all parts collapsed into a single row
• D. 4+2+1 — this partition is self-conjugate

Answer: (B)

The Ferrers diagram of 4+2+1 has rows of length 4, 2, and 1. Reading the column heights left-to-right: column 1 has 3 dots (three rows have at least 1), column 2 has 2 dots (two rows have at least 2), column 3 has 1 dot (one row has at least 3), column 4 has 1 dot (one row has at least 4). This gives 3+2+1+1. Option A is a common confusion between reversing parts and taking the conjugate; the conjugate reads columns, not rows in reverse.

Question 3

The coefficient of x^n in the infinite product ∏(1/(1−x^k)) for k = 1, 2, 3, … equals p(n), the number of partitions of n.

• T. True
• F. False

Answer: True

Each factor 1/(1−x^k) = 1 + x^k + x^(2k) + … encodes how many times part k appears (0, 1, 2, … times). Multiplying over all k = 1, 2, 3, … collects every combination of parts summing to n in the coefficient of x^n. This generating function identity converts the partition-counting problem into a product formula, one of the most powerful tools for studying p(n).

Question 4

The Ramanujan congruences state that p(n) is divisible by 5 for nearly every positive integer n.

• T. True
• F. False

Answer: False

The Ramanujan congruence for the prime 5 states p(5k+4) ≡ 0 (mod 5) — divisibility holds only when n ≡ 4 (mod 5). For example, p(4) = 5, p(9) = 30, p(14) = 135, each divisible by 5; but p(1) = 1 and p(2) = 2 are not. The surprise is that such a clean modular pattern exists at all for a combinatorial quantity — not that every value is divisible.

Question 5

What is the conjugate of a partition, and what theorem does the Ferrers diagram immediately prove about conjugates?

Model answer: The conjugate of a partition is obtained by transposing its Ferrers diagram — reading the column lengths instead of the row lengths. This gives a new partition of the same integer. The Ferrers diagram immediately proves that the number of partitions of n with largest part equal to k equals the number of partitions of n with exactly k parts, because taking the conjugate is a bijection that swaps 'largest part' with 'number of parts.'

The bijective proof via conjugation is a prototype of the combinatorial proof technique: instead of computing two quantities separately and showing they're equal, you exhibit a one-to-one correspondence between the two sets. The Ferrers diagram makes this correspondence visual and concrete — rotating the grid 90° is the bijection.