CombinatoricsTopic #17 of 40

Combinations

Combinations, binomial coefficients, Pascal's triangle, and combinations with repetition.

Definition

A combination is an unordered selection of objects. Order doesn't matter.

Combinations without Repetition

Formula

The number of ways to choose rr objects from nn distinct objects: C(n,r)=(nr)=n!r!(nr)!C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}

Read as "nn choose rr" or "binomial coefficient."

Example: Choose 3 students from 10: (103)=10!3!7!=10×9×83×2×1=120\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

Relationship to Permutations

(nr)=P(n,r)r!\binom{n}{r} = \frac{P(n,r)}{r!}

We divide by r!r! to remove the ordering of selected items.

Properties of Binomial Coefficients

Symmetry

(nr)=(nnr)\binom{n}{r} = \binom{n}{n-r}

Choosing rr to include = choosing nrn-r to exclude.

Pascal's Identity

(nr)=(n1r1)+(n1r)\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}

Either include a specific element (pick r1r-1 more) or don't (pick all rr from rest).

Sum of Row

r=0n(nr)=2n\sum_{r=0}^{n} \binom{n}{r} = 2^n

Total subsets of an nn-set.

Alternating Sum

r=0n(1)r(nr)=0\sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0

Vandermonde's Identity

(m+nr)=k=0r(mk)(nrk)\binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k} \binom{n}{r-k}

Choose rr from m+nm+n by partitioning the selection.

Hockey Stick Identity

i=rn(ir)=(n+1r+1)\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}

Pascal's Triangle

Each entry is the sum of two entries above it.

         1
        1 1
       1 2 1
      1 3 3 1
     1 4 6 4 1
    1 5 10 10 5 1

Row nn contains (n0),(n1),,(nn)\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}.

Combinations with Repetition

When items can be selected more than once (multiset selection):

Formula

Choose rr items from nn types (with unlimited copies): (n+r1r)=(n+r1n1)\binom{n+r-1}{r} = \binom{n+r-1}{n-1}

Stars and Bars interpretation:

  • rr identical stars (items to choose)
  • n1n-1 bars (dividers between types)
  • Arrange rr stars and n1n-1 bars

Example: Buy 5 donuts from 3 flavors: (3+515)=(75)=21\binom{3+5-1}{5} = \binom{7}{5} = 21

Counting Subsets

TypeCount
All subsets of nn-set2n2^n
Subsets of size exactly kk(nk)\binom{n}{k}
Subsets of size at most kki=0k(ni)\sum_{i=0}^{k} \binom{n}{i}
Subsets of size at least kki=kn(ni)\sum_{i=k}^{n} \binom{n}{i}
Non-empty subsets2n12^n - 1

Distributing Objects

Distinct Objects into Distinct Boxes

Each object goes to one of kk boxes: knk^n

Identical Objects into Distinct Boxes

Distribute nn identical balls into kk boxes: (n+k1n)=(n+k1k1)\binom{n+k-1}{n} = \binom{n+k-1}{k-1}

With Minimum Requirements

Each of kk boxes must have at least 1: (n1k1)\binom{n-1}{k-1}

(Place nn balls in a row, choose k1k-1 dividers among n1n-1 gaps)

Committee Problems

Basic Committee

Choose committee of kk from nn people: (nk)\binom{n}{k}

Committee with Chair

Choose committee of kk with a chair: (nk)×k=k(nk)\binom{n}{k} \times k = k \binom{n}{k}

Or: (n1)×(n1k1)=n(n1k1)\binom{n}{1} \times \binom{n-1}{k-1} = n \binom{n-1}{k-1}

Committee from Groups

From 5 men and 7 women, form committee of 4 with at least 2 women:

  • 2 women, 2 men: (72)(52)=21×10=210\binom{7}{2}\binom{5}{2} = 21 \times 10 = 210
  • 3 women, 1 man: (73)(51)=35×5=175\binom{7}{3}\binom{5}{1} = 35 \times 5 = 175
  • 4 women, 0 men: (74)(50)=35×1=35\binom{7}{4}\binom{5}{0} = 35 \times 1 = 35
  • Total: 210+175+35=420210 + 175 + 35 = 420

Lattice Paths

Paths from (0,0)(0,0) to (m,n)(m,n) using only right (R) and up (U) moves: (m+nm)=(m+nn)\binom{m+n}{m} = \binom{m+n}{n}

Choose which mm of the m+nm+n steps are rightward.

Example: Paths from (0,0)(0,0) to (4,3)(4,3): (74)=35\binom{7}{4} = 35