Permutations of like objects
Permutations of like objects is part of combinatorics in Year 11 Specialist Mathematics in Queensland (QCAA). It counts the arrangements of a set when some objects are identical, such as the repeated letters in a word.
You will learn to divide by a factorial for each repeated group and to count shortest paths on a street grid. These skills underpin permutations, combinations and probability later in the course.
Theory
Permutations of like objects counts the arrangements of a set in which some objects are identical, a core skill in Year 11 Specialist Mathematics (QCAA, Queensland). Because swapping two identical objects gives the same arrangement, the plain \(n!\) over-counts; divide by a factorial for each repeated group. This page shows the rule \(\dfrac{n!}{p!\,q!\,r!\cdots}\) applied to words with repeated letters and to shortest grid paths.
A permutation is an ordered arrangement of objects. When every object is different, \(n\) objects arrange in \(n!\) ways. Difficulty appears when some objects are identical (like objects): swapping two identical letters produces the very same arrangement, so \(n!\) counts each real arrangement more than once.
To correct the over-count, divide by the factorial of each repeated group. If a repeated object appears \(p\) times, those \(p\) copies could be ordered among themselves in \(p!\) ways, all giving the same visible arrangement, so dividing by \(p!\) removes the duplicates.
The same idea counts shortest lattice paths on a street grid. A shortest route from one corner to another is just a sequence of unit steps — some East (E) and some North (N) — so counting routes is counting arrangements of a word made of repeated E's and N's.
The key question is: how many of each identical object are there? Each repeated group contributes one factorial in the denominator.
For \(n\) objects in which one kind is repeated \(p\) times, another \(q\) times, another \(r\) times, and so on, the number of distinct arrangements is:
For a shortest path on a grid that needs \(m\) East steps and \(n\) North steps, the moves are an arrangement of \(m+n\) letters (\(m\) E's and \(n\) N's):
Counting arrangements of like objects
- Count the total number of objects, \(n\), and write \(n!\) as the numerator.
- Identify each group of identical objects and how many times it repeats (\(p\), \(q\), \(r\), …).
- Divide \(n!\) by the factorial of each repeated group: \(\dfrac{n!}{p!\,q!\,r!\cdots}\).
- Evaluate the factorials and simplify to a whole number. (For a grid path, first turn the route into E's and N's, then apply the same rule.)
Count the letters, then the repeated groups. BANANA has one B, three A's and two N's:
| \(\text{letters}\) | \(=\) | \(6\) |
| \(\text{repeats}\) | \(:\) | \(\text{A} \to 3,\ \text{N} \to 2\) |
| \(\text{arrangements}\) | \(=\) | \(\dfrac{6!}{3!\,2!}\) |
| \(=\) | \(\dfrac{720}{6 \times 2}\) | |
| \(=\) | \(\dfrac{720}{12}\) | |
| \(=\) | \(60\) |
\(60\) arrangements.
Count the letters, then each repeated group. STATISTICS has three S's, three T's and two I's (one A):
| \(\text{letters}\) | \(=\) | \(10\) |
| \(\text{repeats}\) | \(:\) | \(\text{S} \to 3,\ \text{T} \to 3,\ \text{I} \to 2\) |
| \(\text{arrangements}\) | \(=\) | \(\dfrac{10!}{3!\,3!\,2!}\) |
| \(=\) | \(\dfrac{3\,628\,800}{6 \times 6 \times 2}\) | |
| \(=\) | \(\dfrac{3\,628\,800}{72}\) | |
| \(=\) | \(50\,400\) |
\(50\,400\) arrangements.
Six counters with three identical pairs; divide \(6!\) by \(2!\) for each colour:
| \(\text{counters}\) | \(=\) | \(2+2+2 = 6\) |
| \(\text{repeats}\) | \(:\) | \(\text{G} \to 2,\ \text{Y} \to 2,\ \text{R} \to 2\) |
| \(\text{arrangements}\) | \(=\) | \(\dfrac{6!}{2!\,2!\,2!}\) |
| \(=\) | \(\dfrac{720}{2 \times 2 \times 2}\) | |
| \(=\) | \(\dfrac{720}{8}\) | |
| \(=\) | \(90\) |
\(90\) arrangements.
Turn a route into letters: every shortest route uses exactly 4 E's and 3 N's, so count the arrangements of that 7-letter string:
| \(\text{moves}\) | \(=\) | \(4+3 = 7\) |
| \(\text{repeats}\) | \(:\) | \(\text{E} \to 4,\ \text{N} \to 3\) |
| \(\text{routes}\) | \(=\) | \(\dfrac{7!}{4!\,3!}\) |
| \(=\) | \(\dfrac{5040}{24 \times 6}\) | |
| \(=\) | \(\dfrac{5040}{144}\) | |
| \(=\) | \(35\) |
\(35\) shortest routes.
Common pitfalls
Frequently asked questions
How do you count arrangements when some objects are the same?
Divide the total factorial by the factorial of each repeated group: for \(n\) objects with repeats \(p\), \(q\), \(r\), the count is \(\dfrac{n!}{p!\,q!\,r!\cdots}\).
Why do you divide by a factorial for each repeated letter?
A group of \(p\) identical objects could be reordered among themselves in \(p!\) ways, all giving the same visible arrangement. Dividing by \(p!\) removes those duplicate counts.
How many ways can the letters of BANANA be arranged?
BANANA has three A's and two N's, so the count is \(\dfrac{6!}{3!\,2!}=\dfrac{720}{12}=60\).
How do you count the shortest paths on a grid?
A shortest route that goes \(m\) blocks east and \(n\) blocks north is an arrangement of \(m\) E's and \(n\) N's, so the number of routes is \(\dfrac{(m+n)!}{m!\,n!}\).
What happens to objects that only appear once?
A single object contributes \(1!=1\) to the denominator, which changes nothing. Only groups that repeat two or more times matter.
Is 6!/(3!2!) the same as dividing 6! by 5!?
No. Divide by each group separately and multiply the results: \(3!\times2!=12\), not \(5!=120\). Combining the factorials into one is a common error.