Questions: The Multinomial Theorem and Multinomial Coefficients

Question 1

How many distinct arrangements are there of the letters in the word MISSISSIPPI (11 letters: 1 M, 4 I, 4 S, 2 P)?

• A. 11! = 39,916,800 (treating all letters as distinct)
• B. 11! / (4! × 4!) = 6,930 (forgetting to account for M and P)
• C. 11! / (1! × 4! × 4! × 2!) = 34,650
• D. C(11,4) × C(7,4) = 11,550 (choosing positions for just two letter types)

Answer: (C)

The multinomial coefficient n!/(n₁!n₂!⋯nₖ!) counts arrangements of n objects where n₁ are of type 1, n₂ of type 2, etc. Here n=11, with groups 1M, 4I, 4S, 2P: 11!/(1!×4!×4!×2!) = 39,916,800/1,152 = 34,650. Option A overcounts by treating identical letters as distinct. Option B forgets to divide by 1! and 2! for M and P. Option D only accounts for two letter types instead of all four.

Question 2

The binomial coefficient C(n, k) is a special case of the multinomial coefficient. Under what conditions does the multinomial coefficient reduce to C(n, k)?

• A. When n is even
• B. When all variables in the expansion are set equal to 1
• C. When there are exactly two groups, of sizes k and n−k, so the multinomial coefficient becomes n!/(k!(n−k)!)
• D. When the exponents n₁, n₂, ..., nₖ are all equal

Answer: (C)

C(n,k) = n!/(k!(n−k)!) is exactly the multinomial coefficient for k=2 groups of sizes k and n−k. The multinomial generalizes this to any number of groups. Setting all variables equal to 1 in the multinomial theorem gives the identity that all multinomial coefficients sum to kⁿ — useful, but a different relationship. Equal exponents produce a special class of terms but don't recover the binomial coefficient in general.

Question 3

The multinomial coefficient n!/(n₁!n₂!⋯nₖ!) gives both the number of ways to arrange n objects when nᵢ are of type i AND the coefficient of x₁^n₁ x₂^n₂ ⋯ xₖ^nₖ in the expansion of (x₁ + x₂ + ⋯ + xₖ)^n.

• T. True
• F. False

Answer: True

Both interpretations use the same combinatorial reasoning. When expanding (x₁ + ⋯ + xₖ)^n, a term x₁^n₁ ⋯ xₖ^nₖ arises from choosing which of the n factors contribute each variable — the number of such choices is exactly the multinomial coefficient. This is also the number of ways to arrange n objects when n₁ are indistinguishable type-1, n₂ are indistinguishable type-2, etc. The combinatorial structure is identical.

Question 4

The multinomial theorem only applies when the number of variables equals the exponent — that is, (x₁ + x₂ + ⋯ + xₖ)^n requires k = n.

• T. True
• F. False

Answer: False

There is no requirement that k = n. k is the number of variables (terms in the sum) and n is the exponent; they are independent. For example, (x + y + z)^2 has k=3 variables and n=2, and it expands as x² + y² + z² + 2xy + 2xz + 2yz using multinomial coefficients 2!/(2!0!0!) = 1 for pure squares and 2!/(1!1!0!) = 2 for mixed terms. The constraint is only that n₁ + n₂ + ⋯ + nₖ = n for each term.

Question 5

Explain why the multinomial coefficient n!/(n₁!n₂!⋯nₖ!) correctly counts the number of distinct arrangements of n objects when nᵢ objects of each type i are identical.

Model answer: Start by imagining all n objects are distinct: there are n! total arrangements. But objects of the same type are indistinguishable, so each unique arrangement is overcounted. Specifically, any arrangement is counted n₁! times (because the n₁ identical type-1 objects can be permuted among themselves without creating a new arrangement), and similarly n₂! times for type-2 objects, and so on. Dividing by each nᵢ! removes this overcounting, leaving n!/(n₁!n₂!⋯nₖ!) distinct arrangements.

This is the 'division by overcounting' principle. The anagram formula works by the same logic: MISSISSIPPI has 11! arrangements if all letters are labeled, but swapping the four S's among themselves gives the same word, so we divide by 4!; same for I's (4!) and P's (2!) and M (1!). This principle recurs throughout combinatorics: count all arrangements, identify the symmetries (permutations of identical objects), divide by their count.