Binomial Coefficients and Combinatorial Identities · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

One set, counted two ways

Mathematics · Discrete Mathematics · ages 18-19
Name ______________________   Date ____________
  1. Split 4-person committees from 9 people on whether they include Ana. Which term counts committees that include her?

    • C(8,4)
    • C(9,3)
    • C(8,3)
    • C(9,4)
  2. What is the binomial coefficient C(5, 2)? Type the number.

    Answer: ______________

  3. For integers n >= 0 and 0 <= r <= n, the binomial coefficient C(n,r) is a whole number.

    Circle one:   True   False

  4. What does the binomial coefficient C(n, r) count?

    • The r-person groups from n people
    • The n-person groups from r people
    • The sum of row n of Pascal's triangle
  5. There are C(8,3) = 56 three-person committees from 8 people. Of these, C(7,3) = 35 exclude Ana. How many include Ana?

    Answer: ______________

  6. When Pascal's rule splits pairs from 6 people by whether they include Ana, what are the two cases?

    • Pairs from the first 3 plus pairs from the last 3
    • Pairs including Ana plus pairs excluding her
    • Pairs times singles, then divided by 2
  7. Row 5 of Pascal's triangle reads 1, 5, 10, 10, 5, 1. What do its entries sum to? Type the number.

    Answer: ______________

  8. A proof splits 4-person committees from 9 people into those containing Ana and those containing Bo. Why is this not a valid Pascal partition?

    • Some committees contain both; others contain neither.
    • The two cases have different sizes, so they cannot be added.
    • Every committee must contain Ana, so the Bo case is unnecessary.
    • Fixing a person changes committees into ordered arrangements.
  9. For any integer n >= 0, why does C(n,0) + ... + C(n,n) = 2^n? Choose the counting proof.

    • Both count all subsets: by size on the left and by independent in-or-out decisions on the right.
    • Both count only nonempty subsets: by size on the left and by independent in-or-out decisions on the right.
    • Both count ordered lists: by length on the left and by independent first-or-last placements on the right.
  10. Choose any 3 people from disjoint pools of 4 artists and 5 writers. Which expression counts every team once by its artist count?

    • C(4,0) + C(5,3) + C(4,3) + C(5,0), adding pool counts
    • Sum of C(4,k) × C(5,3-k) for k = 0,1,2,3
    • C(4,1) × C(5,2), counting only one possible team type
    • C(4,3) × C(5,3), choosing three people from each pool
LightMySky · lightmysky.comW1-mt_lHXFvudTh6-s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

One set, counted two ways W1-mt_lHXFvudTh6-s1

  1. C(8,3) · Ana fills one place. Choose the other three members from the remaining eight people: C(8,3).
  2. 10 · C(5,2) = (5 × 4) / (2 × 1) = 20 / 2 = 10.
  3. True · On this integer domain, C(n,r) counts r-element subsets. A count of subsets is a whole number.
  4. The r-person groups from n people · C(n, r) counts groups of size r drawn from n people.
  5. 21 · Including Ana and excluding Ana partition all committees. The missing count is 56-35 = 21, also C(7,2).
  6. Pairs including Ana plus pairs excluding her · Pascal's rule splits the pairs by whether they include the fixed person.
  7. 32 · Row 5 sums to 2 to the power 5, which is 32.
  8. Some committees contain both; others contain neither. · The cases overlap for committees with both people and omit committees with neither. A valid partition must count every committee exactly once.
  9. Both count all subsets: by size on the left and by independent in-or-out decisions on the right. · Each subset has exactly one size, so the row sum counts all subsets once. Independent include-or-exclude decisions also produce each subset once, in 2^n ways. Thus the counts are equal.
  10. Sum of C(4,k) × C(5,3-k) for k = 0,1,2,3 · For each artist count k, choose k artists and 3-k writers. Add the four disjoint case counts: 10 + 40 + 30 + 4 = 84.
Worksheet · LightMySky