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Expanding is choosing
Multiplying out (x+y)ⁿ means picking x or y from each factor. The coefficient of x^{n−k}y^k counts the ways to pick y from exactly k of them.
Lesson · Combinatorics and Counting · Binomial coefficients
The Binomial Theorem
(x + y)ⁿ = Σ C(n,k) x^{n−k} y^k. The coefficient counts the ways to choose k factors to contribute y, which is why the binomial coefficients appear — and the substitutions x = y = 1 recover the row sums.
(x + y)ⁿ = Σ_{k=0}^{n} C(n,k) x^{n−k} y^k
x = y = 1 ⟹ Σ C(n,k) = 2ⁿ
x = 1, y = −1 ⟹ Σ (−1)^k C(n,k) = 0
The coefficient counts CHOICES of which factors supply y.
One theorem, several corollaries
Core ideas
What this lesson establishes
The theorem is a counting statement wearing algebraic notation, and its corollaries come from substituting numbers.
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Substitution generates identities
Setting x and y to specific values turns the theorem into a numerical identity. Two substitutions give the row sum and the alternating sum for free.
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The multinomial theorem generalises it
With more than two terms the coefficients count arrangements of a multiset, which is “Permutations with Repetition”'s subject.
The material
The theorem and what follows
Statement, the counting proof, and four corollaries.
Theorem · Binomial theorem
For any x, y and any n ≥ 0, (x + y)ⁿ = Σ_{k=0}^{n} C(n,k) x^{n−k} y^k.
(x+y)ⁿ is the product of n identical factors (x+y). Expanding without collecting terms produces one term for each way of choosing x or y from each factor — 2ⁿ terms in all.
A term equals x^{n−k}y^k exactly when y was chosen from k of the factors, and the number of ways to choose those k factors is C(n,k).
Collecting equal terms therefore gives coefficient C(n,k) on x^{n−k}y^k.
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This explains the name: the binomial coefficients are the coefficients in a binomial expansion, and they count subsets because expanding is choosing. The two facts are not analogous — they are the same fact.
An induction proof using Pascal's rule is also available and is a good exercise for Induction and Recursion's technique. It verifies the theorem; the counting argument explains it.
Corollaries by substitution
x = 1, y = 1 Σ_k C(n,k) = 2ⁿ
x = 1, y = −1 Σ_k (−1)^k C(n,k) = 0 for n ≥ 1
x = 1, y = 2 Σ_k 2^k C(n,k) = 3ⁿ
x = 1, y = t Σ_k t^k C(n,k) = (1 + t)ⁿ
The last line is the generating function of the binomial coefficients, and “Generating Functions” makes that phrase mean something. For now it is the observation that a single algebraic identity encodes a whole row of the triangle.
Example · Extracting one coefficient
What is the coefficient of x⁵y³ in (2x − 3y)⁸? The general term is C(8,k)(2x)^{8−k}(−3y)^k. Take k = 3: C(8,3)·2⁵·(−3)³ = 56 · 32 · (−27) = −48384.
Two places to be careful: the constants inside the parentheses are raised to the powers too, and a negative sign must be carried through the odd power. Both are dropped often enough to be worth naming.
Theorem · Multinomial theorem
(x₁ + ⋯ + x_m)ⁿ = Σ (n! / (k₁! ⋯ k_m!)) x₁^{k₁} ⋯ x_m^{k_m}, summed over all non-negative kᵢ with k₁ + ⋯ + k_m = n. The coefficient counts the ways to assign the n factors to the m terms, which is “Permutations with Repetition”'s arrangement count.
Interactive
Pascal's triangle
Identities highlighted and checked
Method
Using the theorem
Write the general term first.
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Write the general term
C(n,k) a^{n−k} b^k where a and b are the two whole expressions being added, constants and signs included.
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Solve for k from the power you want
Match the exponent of one variable and read off k. Matching both is a useful check that the term exists.
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Evaluate the constants carefully
(2x)⁵ contributes 2⁵, and (−3y)³ contributes (−3)³ = −27. Signs on odd powers are the standard slip.
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Substitute to get identities
Setting x and y to 1, −1 or small integers turns the theorem into numerical facts about a row of the triangle.
The binomial theorem
Score 0 / 0Choose the answer you can justify.
Expand and extract
Two uses of the same general term.
(x + y)⁴ = C(4,0)x⁴ + C(4,1)x³y + C(4,2)x²y² + C(4,3)xy³ + C(4,4)y⁴
= x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴
coefficients 1, 4, 6, 4, 1 — row 4 of Pascal's triangle
Coefficient of x²y⁶ in (x + 2y)⁸:
general term C(8,k) x^{8−k} (2y)^k
want y⁶ ⟹ k = 6, and then x^{8−6} = x² ✓
C(8,6)·2⁶ = 28 · 64 = 1792
The check that x^{8−k} gives the exponent you wanted is worth the second it takes. If the two exponents do not both match, the term you are looking for does not appear in the expansion at all.
Common mistakes
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Forgetting the constants inside the parentheses(2x)⁵ is 32x⁵. The coefficient of the term is C(n,k) times whatever the constants contribute.
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Losing a sign on an odd power(−y)³ = −y³. Half of all extraction errors are this one.
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Confusing the index with the exponentIn C(n,k)x^{n−k}y^k, k is the exponent of the SECOND term. Reading it as the first gives the mirror-image coefficient, which is right only when the row is symmetric at that point.
Check your understanding
✓
Finish when you can extract a coefficient with constants and signs present.
Find the coefficient of x⁴y⁵ in (2x − 3y)⁹, and check the sign and both constants. Then set x = y = 1 in the same expansion and confirm the result equals (2 − 3)⁹ = −1.
Note: Newton extended the theorem to non-integer exponents, where the sum becomes an infinite series and the coefficients use the generalised binomial coefficient. That is where “Generating Functions”'s generating functions acquire their power, and it is outside this course.