Questions: Binomial Coefficients and Pascal's Triangle

Question 1

Pascal's identity states that C(5,2) = C(4,1) + C(4,2). Which explanation best captures why this is true?

• A. It is a coincidence that holds for small values but breaks down for large n
• B. Every 2-element selection from 5 items either includes a designated special item (needing C(4,1) additional choices) or excludes it (needing C(4,2) choices from the rest)
• C. Pascal's triangle is defined by adding adjacent entries as an arithmetic rule, with no deeper meaning
• D. The factorial formula simplifies algebraically to the sum of the two smaller values

Answer: (B)

This is the combinatorial proof of Pascal's identity. Designate any one item as 'special.' Every 2-item selection either includes or excludes it. If it includes the special item, you choose 1 more from 4: C(4,1) ways. If it excludes it, you choose 2 from 4: C(4,2) ways. These cases are mutually exclusive and exhaustive, so they sum to C(5,2). Option D is true but misses the point — the algebraic fact follows from combinatorial logic, not the other way around.

Question 2

A student expanding (a+b)⁴ needs the coefficient of a²b². They identify it as C(4,2) = 6. What is the best explanation for why C(4,2) is correct?

• A. It is the entry in row 4, position 2 of Pascal's triangle — which is true by definition of the triangle
• B. Expanding (a+b)⁴ means choosing a or b from each of 4 factors; the term a²b² requires choosing b from exactly 2 of the 4 factors, and there are C(4,2) ways to make that choice
• C. The coefficients must be symmetric and peak in the middle, so the middle terms of a degree-4 expansion are automatically largest
• D. The formula 4!/(2!2!) = 6 holds by algebraic necessity, with no combinatorial interpretation

Answer: (B)

The binomial theorem connects combination-counting to polynomial expansion. Expanding (a+b)⁴ = (a+b)(a+b)(a+b)(a+b) means choosing a or b from each factor independently. To produce a²b², you need to choose b from exactly 2 of the 4 factors. The number of ways to pick which 2 factors contribute b is — by definition — C(4,2) = 6. This is the same counting argument that defines C(n,k) in the first place.

Question 3

The sum of all entries in row n of Pascal's triangle equals 2^n, because it counts all subsets of an n-element set.

• T. True
• F. False

Answer: True

The entries in row n are C(n,0) + C(n,1) + ... + C(n,n). Each C(n,k) counts the k-element subsets. Summing over all k gives the total number of subsets of an n-element set, which is 2^n — each element is either included or excluded, giving 2 independent binary choices for n items. The explainer notes this explicitly as one of Pascal's triangle's deeper patterns, each of which reflects a combinatorial identity.

Question 4

Pascal's identity C(n,k) = C(n-1,k-1) + C(n-1,k) can primarily be verified by algebraic manipulation of the factorial formula — it has no intuitive combinatorial explanation.

• T. True
• F. False

Answer: False

Pascal's identity has a direct combinatorial proof: designate any one object as 'special.' Any k-element selection either includes the special object — leaving C(n-1,k-1) ways to complete it — or excludes it, leaving C(n-1,k) ways to choose all k from the remaining n-1. These cases are mutually exclusive and exhaustive, so they sum to C(n,k). The arithmetic addition rule in Pascal's triangle encodes this logical split. Algebraic verification follows from the combinatorial reality, not the reverse.

Question 5

Why do binomial coefficients appear both as entries in Pascal's triangle and as the coefficients in the expansion of (a+b)^n?

Model answer: Because both contexts are counting the same thing: the number of ways to choose k items from n. In Pascal's triangle, C(n,k) is defined as the number of k-element selections from n objects. In the binomial expansion, C(n,k) appears as the coefficient of a^(n-k)b^k because the term a^(n-k)b^k arises when b is chosen from exactly k of the n factors of (a+b)^n — and the number of ways to choose which k factors contribute b is, by definition, C(n,k).

The triangle and the expansion are not two separate facts that happen to share the same numbers — they reflect the same combinatorial structure. Recognizing binomial coefficients in new settings (probability, polynomial algebra, combinatorial identities) is a core skill in discrete mathematics precisely because the underlying counting argument is the same in all of them.