Questions: Binomial Theorem and Binomial Coefficients

Question 1

In expanding (x + y)^5, why does the term x^3y^2 have coefficient C(5,2) = 10?

• A. Because 5 × 4 / 2! = 10 counts the arrangements of 3 x's and 2 y's in a string of length 5
• B. Because each term comes from choosing which 2 of the 5 factors contribute y, and there are C(5,2) = 10 such choices
• C. Because degree-5 terms always appear with coefficient 10 by the symmetry of Pascal's triangle
• D. Because the Pascal identity gives C(5,2) = C(4,1) + C(4,2) = 4 + 6 = 10 without needing a deeper reason

Answer: (B)

When you multiply five factors of (x+y), each term arises by choosing x or y from each factor independently. To get x^3y^2, you pick y from exactly 2 of the 5 factors — choosing which 2 factors contribute the y. There are C(5,2) = 10 ways to choose that 2-element subset. Option A gives the right number but as a formula, not the combinatorial reason. Option D is valid arithmetic but doesn't explain why that coefficient appears in the expansion.

Question 2

Setting x = y = 1 in the binomial theorem produces an identity. What does that identity count, combinatorially?

• A. The total number of permutations of an n-element set, which is n!
• B. The total number of subsets of an n-element set, which is 2^n — each subset corresponds to choosing which elements to include
• C. The number of ways to choose 2 elements from an n-element set, which is C(n,2)
• D. The sum of the first n natural numbers, which is n(n+1)/2

Answer: (B)

Setting x = y = 1 gives (1+1)^n = 2^n = Σ C(n,k). So the sum of all binomial coefficients in row n is 2^n. Combinatorially, each C(n,k) counts the k-element subsets of an n-element set, and summing over all k counts every subset of every size — there are exactly 2^n subsets total, one for each binary include/exclude decision per element.

Question 3

The coefficient of x^(n-k)y^k in the expansion of (x + y)^n equals the number of k-element subsets of an n-element set.

• T. True
• F. False

Answer: True

This is the combinatorial heart of the binomial theorem. Each term in the expansion comes from choosing x or y from each of the n factors. The x^(n-k)y^k term arises by choosing y from exactly k of the n factors — a k-element subset of the n factors. There are C(n,k) such subsets, so C(n,k) is the coefficient. The algebra and the combinatorics describe the same counting problem.

Question 4

The binomial coefficients are asymmetric: in general, C(n,k) ≠ C(n, n−k).

• T. True
• F. False

Answer: False

C(n,k) = C(n, n−k) always — the symmetry visible in every palindromic row of Pascal's triangle. Choosing k elements to include in a subset is equivalent to choosing the n−k elements to exclude; both choices define the same subset. So C(8,3) = C(8,5) = 56. Students who forget this symmetry may waste computation, and it also underlies the fact that Pascal's triangle reads the same forwards and backwards.

Question 5

Why does substituting x = 1 and y = −1 into the binomial theorem show that any nonempty set has equally many even-sized and odd-sized subsets?

Model answer: Substituting gives (1−1)^n = 0 = Σ (−1)^k C(n,k) = C(n,0) − C(n,1) + C(n,2) − .... Rearranging: C(n,0) + C(n,2) + ... = C(n,1) + C(n,3) + ..., meaning the sum of even-indexed coefficients equals the sum of odd-indexed ones. Since the total is 2^n, each half equals 2^(n−1) — exactly half the subsets have even size and half have odd size.

This substitution technique extracts combinatorial information from algebraic identities. The 0 on the left forces positive and negative terms to cancel exactly — a non-obvious result that becomes transparent once written as a polynomial identity. The technique generalizes: plugging in roots of unity extracts residue-class information about coefficients, which is a core tool in generating functions and inclusion-exclusion.