Vandermonde's identity states C(m+n, r) = Σₖ C(m,k)·C(n,r−k). Which reasoning best justifies why this equality holds?
AThe left side and right side are both polynomials in m and n that happen to be equal by coincidence
BBoth sides count the same thing: the number of ways to choose r items from two groups of m and n by partitioning the selection between the groups in all possible ways
CThe identity follows from applying Pascal's rule repeatedly to the left side
DBinomial coefficients are symmetric, so the product of two of them always simplifies
The combinatorial argument is the heart of the identity: choosing r items from m+n is equivalent to choosing k from the first group and r−k from the second, for every valid k, and summing. Both sides count exactly the same selections. This two-group counting argument is more illuminating than algebraic verification because it explains *why* the identity is true.
Question 2 Multiple Choice
You need to compute Σₖ₌₀^{8} C(8,k)². Which value does Vandermonde's identity imply this equals?
AC(16,8)
B2^8
CC(8,4)²
DC(16,4)
Setting m = n = r = 8 in Vandermonde's identity gives Σ C(8,k)·C(8,8−k) = C(16,8). Since C(8,8−k) = C(8,k) by symmetry of binomial coefficients, this becomes Σ C(8,k)² = C(16,8). This is the special case C(2n,n) = Σₖ C(n,k)² — a striking result most easily seen through the two-group counting argument.
Question 3 True / False
When computing Σₖ C(m,k)·C(n,r−k), you should carefully identify which values of k give nonzero terms before summing, since out-of-range terms could corrupt the result.
TTrue
FFalse
Answer: False
C(m,k) = 0 whenever k < 0 or k > m, and C(n,r−k) = 0 whenever r−k < 0 or r−k > n. Out-of-range terms automatically contribute 0 to the sum. You can let k run from 0 to r (or even over all integers) without tracking bounds — the out-of-range cases vanish on their own. This is one of the practical conveniences of working with binomial coefficients.
Question 4 True / False
Vandermonde's identity can be derived by multiplying the generating functions (1+x)^m and (1+x)^n and comparing coefficients of x^r on both sides.
TTrue
FFalse
Answer: True
(1+x)^m · (1+x)^n = (1+x)^{m+n}. The coefficient of x^r on the right is C(m+n,r). On the left, multiplying the two power series gives Σₖ C(m,k)·C(n,r−k) as the coefficient of x^r. So the two expressions must be equal — this is the generating function proof. It connects the identity to a broader algebraic framework beyond just combinatorial counting.
Question 5 Short Answer
Explain in your own words why the combinatorial proof of Vandermonde's identity is more illuminating than algebraically verifying that both sides are equal.
Think about your answer, then reveal below.
Model answer: The combinatorial proof reveals *why* the identity is true: both sides count the same thing — the ways to choose r items from two groups — through two different but equivalent methods. Algebraic verification confirms the formula is correct but doesn't explain the underlying structure. Understanding the combinatorial reason also extends your ability to recognize similar two-group counting arguments in new problems and to generalize the identity.
A proof that explains 'why' builds intuition and transferable reasoning; a proof that only confirms 'that' leaves the result as a formula to memorize. The combinatorial argument is the insight; the algebra is just a check.