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Evaluate factorial and combination expressions

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Factorial and combination values

Use non-negative integer factorials. For an unordered selection of distinct objects without replacement, divide the ordered count by the factorial of the selection size; the selection size must be between zero and the total number of objects, inclusive

\[\binom{n}{r}=\frac{n!}{r!\left(n-r\right)!}\]
  1. 1Use the factorial definition For a non-negative integer n, n! is the product of the positive integers from n down to 1; define 0!=1. Negative and non-integer factorials are not used here \(\displaystyle 0!=1,\quad4!=4\times3\times2\times1\)
  2. 2Cancel consecutive factorial factors For integers n≥m≥0, expand only the factors needed to cancel m! from n!. Equal numerator and denominator factorials give 1. \(\displaystyle \frac{8!}{5!}=8\times7\times6\)
  3. 3Use combinations for unordered groups Select r different objects from n distinct objects without replacement, with integers 0≤r≤n. Divide the ordered count by r! because rearranging a selected group does not change it \(\displaystyle \binom{8}{3}=\frac{8!}{3!5!}\)
  4. 4Keep order only when positions matter For r different objects placed in r labelled positions, order matters. Use n!/(n−r)! for integers 0≤r≤n, without dividing by r! \(\displaystyle P\left(8,3\right)=8\times7\times6\)
  5. 5Reserve a required object If one specified object must be in an unordered group of r, include it first and choose the other r−1 from the remaining n−1. This requires 1≤r≤n. \(\displaystyle \binom{n-1}{r-1}\)
Year 11, VCE Units 1 & 2, Chapter 2, 4 difficulty levels, No calculatorBuild a worksheet

Factorials and combinations

Factorials multiply consecutive positive integers. Combinations count unordered selections, so repeated orders are divided out

\[\binom{n}{r}=\frac{n!}{r!\left(n-r\right)!}\]
  1. Use the factorial definition For a non-negative integer n, n! is the product of the positive integers from n down to 1; define 0!=1. Negative and non-integer factorials are not used here \(\displaystyle 0!=1,\quad4!=4\times3\times2\times1\)
  2. Cancel consecutive factorial factors For integers n≥m≥0, expand only the factors needed to cancel m! from n!. Equal numerator and denominator factorials give 1. \(\displaystyle \frac{8!}{5!}=8\times7\times6\)
  3. Use combinations for unordered groups Select r different objects from n distinct objects without replacement, with integers 0≤r≤n. Divide the ordered count by r! because rearranging a selected group does not change it \(\displaystyle \binom{8}{3}=\frac{8!}{3!5!}\)
  4. Keep order only when positions matter For r different objects placed in r labelled positions, order matters. Use n!/(n−r)! for integers 0≤r≤n, without dividing by r! \(\displaystyle P\left(8,3\right)=8\times7\times6\)
  5. Reserve a required object If one specified object must be in an unordered group of r, include it first and choose the other r−1 from the remaining n−1. This requires 1≤r≤n. \(\displaystyle \binom{n-1}{r-1}\)

Worked example

Evaluate the combination below

\[\binom{9}{2}\]
  1. \(\displaystyle 9-2=7\)The number of unselected objects is 7
  2. \(\displaystyle \binom{9}{2}=\frac{9!}{2!7!}=\frac{9\times8}{2\times1}\)Use the combination formula and cancel the common factorial factor
  3. \(\displaystyle \frac{9\times8}{2\times1}=36\)Evaluate the remaining product to obtain 36

Common mistakes

Treating a combination as ordered
✕\(\displaystyle \binom{8}{3}=8\times7\times6=336\)✓\(\displaystyle \binom{8}{3}=\frac{8\times7\times6}{3\times2\times1}=56\)
A combination counts one selected group, so repeated orders must be divided out
Distributing a factorial over a sum
✕\(\displaystyle \left(5+2\right)!=5!+2!=122\)✓\(\displaystyle \left(5+2\right)!=7!=5040\)
The factorial applies to the total sum before the factorial is evaluated
Using an incomplete combination denominator
✕\(\displaystyle \binom{9}{3}=\frac{9!}{3!\times6}=10080\)✓\(\displaystyle \binom{9}{3}=\frac{9!}{3!6!}=84\)
The denominator contains the factorial of both the selected count and the unselected count

Difficulty levels

introEvaluate small factorials and direct combinations
standardCancel factorial ratios and count unordered selections
challengeEvaluate composed ratios and selections with a required object
examDistinguish ordered from unordered selections and evaluate the count

Questions about this topic

Why is the factorial of zero equal to one?+

It is the multiplicative identity required by the factorial recurrence and by the combination formula

When is a combination used instead of an arrangement?+

Use a combination when changing the order of the selected objects does not create a new outcome

How can a factorial ratio be evaluated efficiently?+

Expand only the consecutive factors that do not cancel, then simplify before multiplying

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 2: Algebraic foundations

In this chapter

Expand binomial and special productsFactorise by common factors and difference of perfect squaresFactorise monic and non-monic quadratic trinomialsFactorise four-term expressions by groupingRecognise equivalent polynomial formsSimplify and operate on algebraic fractionsSimplify and combine algebraic fractionsEvaluate factorial and combination expressionsPascal rows and binomial coefficient structureLocate and evaluate a term in a binomial expansionClassify real numbers and recognise undefined formsWork with sets and interval notationSimplify, compare and operate with surdsRationalise a surd denominator