1
Drawing with replacement gives nʳ
Each of the r positions is filled independently from all n options. This is the product rule with no shrinkage.
Lesson · Combinatorics and Counting · Repetition
Permutations with Repetition
Two different situations share this name. Choosing r items from n with repetition allowed and order mattering gives nʳ. Arranging a multiset divides by the factorials of the repetition counts.
ordered, repetition allowed: nʳ
arrangements of a multiset: n! / (n₁! n₂! ⋯ n_k!)
MISSISSIPPI: 11! / (1!·4!·4!·2!) = 34 650
The divisor is the symmetry you are quotienting out.
Two counts, one name
Core ideas
What this lesson establishes
Repetition appears in two guises, and they need different formulas. Distinguishing them is the lesson.
2
Identical objects divide out
Arranging n objects of which some are indistinguishable gives n! divided by the factorial of each repetition count.
3
The divisor is a symmetry group size
Swapping two identical letters produces the same word, so each distinct word was counted once for every such swap — exactly the circular-arrangement argument of “Permutations”.
The material
Repetition in arrangements
Both counts, derived, and the connection to binomial coefficients.
Theorem · Ordered selections with repetition
The number of sequences of length r from an n-element set, with repetition allowed, is nʳ.
Each position is filled independently, with all n options available every time. The product rule gives n · n ⋯ n = nʳ.
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This is the count of functions from an r-set to an n-set, and of strings of length r over an alphabet of size n. {0,1}ⁿ has 2ⁿ elements for exactly this reason.
Theorem · Arrangements of a multiset
If n objects consist of k distinct types with n₁, …, n_k copies (n₁ + ⋯ + n_k = n), the number of distinguishable arrangements is n! / (n₁! n₂! ⋯ n_k!).
Label the identical copies temporarily so all n objects are distinct; there are n! arrangements. Each distinguishable arrangement arises from exactly n₁! n₂! ⋯ n_k! of them, one for each way of permuting the labels within each type. Dividing gives the count.
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Example · MISSISSIPPI
Eleven letters: one M, four I, four S, two P. The number of distinguishable arrangements is 11!/(1!·4!·4!·2!) = 39 916 800 / 1152 = 34 650, rather than 11! = 39 916 800.
The connection to binomial coefficients
C(n, r) is a special case: arranging r copies of "in" and n − r copies of "out" gives n!/(r!(n−r)!), which is exactly C(n,r). Choosing a subset and arranging a two-type multiset are the same problem.
Theorem · Multinomial coefficient
n! / (n₁! ⋯ n_k!) is written binom(n; n₁, …, n_k) and counts both the arrangements of the multiset and the ways to distribute n distinct objects into k labelled boxes with prescribed sizes. It is the coefficient in the multinomial theorem of “The Binomial Theorem”.
Example · Distributing into boxes
Deal 52 cards to four players, 13 each: 52!/(13!)⁴ ≈ 5.36 × 10²⁸. The formula is the same because assigning each card a player is arranging a multiset of 13 copies of each of four labels.
Lattice paths again
A path from (0,0) to (m,n) using unit right and up steps is an arrangement of m R's and n U's, so there are (m+n)!/(m! n!) = C(m+n, m) of them — the same answer “Combinations” gave, now visible as a multiset arrangement.
Interactive
The four counting rules
Enumerated while enumeration is possible
Method
Counting with repetition
Decide which of the two situations you are in.
1
Are you drawing with replacement, or arranging a fixed multiset?
Drawing gives nʳ; arranging gives the factorial quotient. They are different questions with different answers.
2
For drawing, apply the product rule
n options at every one of the r positions, with no shrinkage.
3
For arranging, count the repetitions of each type
The divisor is the product of the factorials of those counts. Missing a repeated letter inflates the answer.
4
Check the exponents sum to n
n₁ + ⋯ + n_k must equal the total number of objects. If it does not, a type has been miscounted.
Repetition
Score 0 / 0Choose the answer you can justify.
Arrangements with a constraint
How many arrangements of BANANA have no two A's adjacent?
BANANA: 6 letters — 3 A, 2 N, 1 B.
Total arrangements: 6!/(3!·2!·1!) = 720/12 = 60
For no two A's adjacent, place the non-A letters first:
arrange B, N, N: 3!/2! = 3 ways e.g. B N N
this creates 4 gaps: _ B _ N _ N _
choose 3 of the 4 gaps for the A's: C(4,3) = 4
3 × 4 = 12 arrangements with no two A's adjacent
So 60 − 12 = 48 arrangements DO have two A's adjacent.
The gap method is the standard device for "no two adjacent" constraints: arrange the unconstrained objects, then insert the constrained ones into the gaps between them. It works because choosing distinct gaps guarantees no two of the inserted objects touch.
Common mistakes
1
Confusing the two situations"Repetition allowed" when drawing gives nʳ; "repeated objects" when arranging gives a quotient. They are different problems.
2
Missing a repeated letterEvery repetition contributes a factorial to the divisor. Count the occurrences of each letter explicitly before dividing.
3
Applying the gap method without checking the gap countArranging m objects creates m + 1 gaps, including the ends. Using m − 1 counts only the internal gaps and undercounts.
Check your understanding
✓
Finish when you can handle an adjacency constraint by gaps.
Count the arrangements of MISSISSIPPI with no two S's adjacent. Arrange the other seven letters, count the gaps, and choose four of them. Both halves — the multiset arrangement and the gap choice — are this lesson.
Note: The multinomial coefficient counts three things at once: arrangements of a multiset, distributions of distinct objects into labelled boxes of given sizes, and coefficients in the multinomial expansion. Recognising a problem as any one of them gives you the other two.