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Overview

Many counting problems ask us to split objects into groups.

For example:

  • Divide 1010 students into two teams of 55
  • Split 1212 people into groups of sizes 3,4,53,4,5
  • Separate objects into groups where the groups may or may not be labelled
  • Divide objects with repeated types into smaller collections

The main challenge is deciding whether the groups are labelled.

If the groups have names, such as Team A and Team B, then the groups are labelled.

If the groups have no names, then the groups are unlabelled, and we have to avoid counting the same grouping multiple times.

Motivation

Suppose 88 students are split into two teams of 44.

If the teams are called Red Team and Blue Team, then choosing the Red Team automatically determines the Blue Team.

The number of ways is

(84).\binom{8}{4}.

However, if the two teams are not named, then choosing one group of 44 first and the other group second counts every split twice.

So the number of unlabelled splits is

12(84).\frac{1}{2}\binom{8}{4}.

This difference between labelled and unlabelled groups is one of the most common sources of mistakes in combinatorics.

Labelled Groups

Suppose we have nn distinct objects and want to split them into labelled groups of sizes

n1,n2,…,nk,n_1,n_2,\ldots,n_k,

where

n1+n2+⋯+nk=n.n_1+n_2+\cdots+n_k=n.

The number of ways is

n!n1!n2!⋯nk!.\frac{n!}{n_1!n_2!\cdots n_k!}.

This is because we choose which objects go into the first group, then the second group, and so on.

Example 1

How many ways are there to split 1010 students into three labelled groups of sizes 2,3,52,3,5?

The groups are labelled, so the answer is

10!2!3!5!.\frac{10!}{2!3!5!}.

Equivalently, we can choose:

(102)\binom{10}{2}

students for the first group, then

(83)\binom{8}{3}

students for the second group, and the remaining 55 students form the third group.

As such, the answer is

(102)(83)=10!2!3!5!.\binom{10}{2}\binom{8}{3} = \frac{10!}{2!3!5!}.

Unlabelled Groups of Different Sizes

If the group sizes are all different, then the sizes themselves identify the groups.

For example, if we split 99 people into groups of sizes 2,3,42,3,4, then the group of size 22 cannot be confused with the group of size 33.

So the number of ways is still

9!2!3!4!.\frac{9!}{2!3!4!}.

No extra division is needed.

Example 2

How many ways are there to split 99 students into unlabelled groups of sizes 2,3,42,3,4?

Since the group sizes are different, the answer is

9!2!3!4!.\frac{9!}{2!3!4!}.

Unlabelled Groups of Equal Sizes

If some groups have the same size and are unlabelled, then we must divide by the number of ways to reorder those equal sized groups.

Example 3

How many ways are there to split 88 students into two unlabelled groups of 44?

First choose 44 students for one group:

(84).\binom{8}{4}.

But this counts each split twice, because choosing the first group or choosing its complement gives the same final split.

So the answer is

12(84)=35.\frac{1}{2}\binom{8}{4}=35.

Example 4

How many ways are there to split 1212 students into three unlabelled groups of 44?

If the groups were labelled, the answer would be

12!4!4!4!.\frac{12!}{4!4!4!}.

But the three groups are unlabelled, so we divide by

3!.3!.

So the answer is

12!4!4!4!3!.\frac{12!}{4!4!4!3!}.

Mixed Group Sizes

Sometimes only some groups have equal sizes.

Example 5

How many ways are there to split 1111 students into unlabelled groups of sizes 2,2,3,42,2,3,4?

If the groups were labelled, the answer would be

11!2!2!3!4!.\frac{11!}{2!2!3!4!}.

The two groups of size 22 are indistinguishable, so we divide by

2!.2!.

The groups of size 33 and 44 are unique by size, so no further division is needed.

So the answer is

11!2!2!3!4!2!.\frac{11!}{2!2!3!4!2!}.

Identical Objects and Groups

If the objects are identical and the groups are unlabelled, the problem becomes a partition problem.

For example, splitting 77 identical balls into 33 unlabelled nonempty groups means finding unordered positive integer triples whose sum is 77.

The possible group sizes are

1,1,51,1,5
1,2,41,2,4
1,3,31,3,3
2,2,3.2,2,3.

So there are 44 ways.

Unlike labelled distribution, there is no simple single formula for every unlabelled identical object grouping problem. These are often handled using casework.

Mixtures of Object Types

Some problems involve repeated types of objects.

For example, suppose we have:

  • 33 red balls
  • 22 blue balls
  • 11 green ball

If we want to split them into groups, we must track how many objects of each type go into each group.

These problems are usually solved by:

  1. Deciding the possible type counts in each group
  2. Counting valid cases
  3. Dividing by symmetries if the groups are unlabelled

The key is to avoid treating identical objects as distinct.

Worked Examples

Example 6

How many ways are there to split 66 students into three unlabelled pairs?

If the pairs were labelled, the answer would be

6!2!2!2!.\frac{6!}{2!2!2!}.

But the three pairs are unlabelled, so divide by

3!.3!.

The answer is

6!2!2!2!3!=15.\frac{6!}{2!2!2!3!}=15.

Example 7

How many ways are there to split 1010 students into two unlabelled groups of 33 and one group of 44?

If the groups were labelled, the answer would be

10!3!3!4!.\frac{10!}{3!3!4!}.

The two groups of size 33 are indistinguishable, so divide by

2!.2!.

So the answer is

10!3!3!4!2!.\frac{10!}{3!3!4!2!}.

Example 8

How many ways are there to split 55 identical balls into two unlabelled non empty groups?

We need unordered positive pairs adding to 55.

The possibilities are

1+41+4

and

2+3.2+3.

So there are 22 ways.

Key Contest Insight

When splitting into groups, first decide whether the groups are labelled.

If the groups are labelled, choose objects for each group directly.

If the groups are unlabelled, count as though they were labelled and then divide by the number of ways to permute identical groups.

The most common mistake is forgetting to divide when groups of the same size are unlabelled.

Strategy Tips

  • Decide whether the objects are distinct or identical
  • Decide whether the groups are labelled or unlabelled
  • For distinct objects and labelled groups, use factorial division
  • For unlabelled equal sized groups, divide by the factorial of the number of identical groups
  • If group sizes are all different, no extra division is needed
  • For identical objects into unlabelled groups, use casework or partitions
  • Always check whether two counted outcomes are actually the same grouping

Common Pitfalls

  • Treating unlabelled groups as labelled
  • Dividing by too much when group sizes are different
  • Forgetting to divide by 3!3! when forming three identical groups
  • Treating identical objects as distinct
  • Confusing grouping with distribution to named people
  • Missing cases when identical objects are split into unlabelled groups

Practice Problems

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