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Worked solution

Partition children into unlabelled groups

In how many ways can a party of 10 children be divided into

(i) one group of 2 and one group of 8,

(ii) two groups of 5,

(iii) three groups of sizes 2,4 and 4?

CJC · Lecture · Example 10

(i) Groups of different sizes identify themselves

Choose the 2-person group, then all 8 remaining children form the other group: (102)(88)=45×1=45\binom{10}{2}\binom88=45\times1=45. The different group sizes prevent a stage swap from creating a second representation of this same selection process; do not divide by 2!.

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(ii) Divide out the two orders of the equal groups

Choosing a first 5-person set and then its remaining complement gives (105)(55)=252×1\binom{10}{5}\binom55=252\times1. But selecting C₁,…,C₅first and C₆,…,C₁₀second describes the same two unlabelled groups as selecting them in the opposite order. The source drawing shows these two stage orders. Divide by 2!: (105)(55)/2!=252/2=126\binom{10}{5}\binom55/2!=252/2=126.

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(iii) Only the two four-person groups can be interchanged

Choose 2 children from 10, then 4 from the remaining 8, then the last 4. The ordered-stage count is (102)(84)(44)=45×70×1\binom{10}{2}\binom84\binom44=45\times70\times1. The two 4-person groups are indistinguishable, so divide by 2!: the answer is 45×70/2=157545\times70/2=1575. Do not divide by 3!: the 2-person group is distinguished by its size.

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The source double-counting illustration: first select C1,C2,C3,C4,C5 then C6,C7,C8,C9,C10; swapping which set is Stage 1 and Stage 2 gives exactly the same two unlabelled groups. Two stage orders correspond to one partition, so divide by 2 factorial.
The source double-counting illustration: first select C1,C2,C3,C4,C5 then C6,C7,C8,C9,C10; swapping which set is Stage 1 and Stage 2 gives exactly the same two unlabelled groups. Two stage orders correspond to one partition, so divide by 2 factorial.
The source double-counting illustration: first select C1,C2,C3,C4,C5 then C6,C7,C8,C9,C10; swapping which set is Stage 1 and Stage 2 gives exactly the same two unlabelled groups. Two stage orders correspond to one partition, so divide by 2 factorial.

Common mistakes

Divide only for permutations of unlabelled groups that represent the same partition. Equal-sized groups with different assigned names or functions would remain distinguishable.

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