Multinomial coefficients & repeated items
Arranging n items where the items come in groups of identical copies gives n!/(n₁!n₂!…n_k!) — the multinomial coefficient. It generalises C(n,k), which is the two-group case, and it also counts the ways to split n distinct objects into labelled groups of the given sizes. Dividing by the factorial of each repeat count is what removes indistinguishable rearrangements.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Count the total items and each repeatTally how many times each distinct symbol appears. The counts must sum to n.
- Apply the formulan! divided by the product of the factorials of the repeat counts.
- Handle restrictions by blockingTo keep certain letters together, glue them into one unit and arrange the units, then arrange within the block.
- Cross-check with the binomial caseWith two groups the formula reduces to C(n, n₁), which is a useful sanity check.
Worked example
How many distinct arrangements are there of the letters in MISSISSIPPI?
- 11 letters total: M×1, I×4, S×4, P×2.
- Check: 1 + 4 + 4 + 2 = 11 ✓.
- Arrangements: 11!/(1!·4!·4!·2!).
- = 39,916,800/(1 × 24 × 24 × 2) = 39,916,800/1152.
Answer. 34,650 distinct arrangements.
Where marks get dropped
These are the specific errors that cost credit on multinomial coefficients & repeated items questions — QED's rubric penalises each of them separately.
- Dividing by the number of distinct letters instead of the factorials of their counts.
- Forgetting a repeated letter entirely. Every symbol needs its own factorial in the denominator, even the singletons (which contribute 1!).
- Treating identical items as distinguishable, which multiplies the answer by all those factorials.
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Multinomial coefficients & repeated items — frequently asked questions
How does this relate to C(n,k)?
C(n,k) = n!/(k!(n−k)!) is the two-group multinomial: choose which positions hold the first symbol and the rest hold the second.
What is the multinomial theorem?
(x₁+…+x_k)ⁿ expands with coefficients n!/(n₁!…n_k!) on each monomial x₁^n₁…x_k^n_k, exactly generalising the binomial theorem.
Does it count distributions too?
Yes — splitting n distinct objects into labelled boxes of prescribed sizes uses the same formula, since the order within each box does not matter.
The rest of Combinatorics
Counting principles, permutations, combinations. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Sum & product rules
- 2Permutations & combinations
- 3Binomial theorem & Pascal’s triangle
- 4The pigeonhole principle
- 5Inclusion–exclusion
- 6Counting with repetition (stars & bars)
- 7Derangements & counting surjections
- 8Double counting & bijective proofs
- 9Setting up & solving counting recurrences
- 10Generating functions — an introduction
- 11Multinomial coefficients & repeated items
- 12Hockey stick & Vandermonde identities
- 13Catalan numbers & lattice-path counting
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