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Worked solution

Divide ten people into repeated group sizes

Find the number of ways in which 10 people may be divided into three groups of 1, two groups of 2 and one group of 3.

CJC · Lecture · Self-Practice 3

Separate internal order from order of equal-sized groups

The group sizes are 1,1,1,2,2,3, summing to 10. First count assignments into these six temporarily labelled groups: 10!/[(1!)3(2!)2 3!]10!/[(1!)^3(2!)^2\,3!]. Internal permutations within a pair or triple do not create another group.

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Remove the labels of interchangeable groups

The three singleton groups can be permuted in 3! ways, and the two pairs in 2! ways, without changing the partition. Hence the count is 10!/[(1!)3(2!)2 3! 3! 2!]=12,60010!/[(1!)^3(2!)^2\,3!\,3!\,2!]=12{,}600, matching the printed 12600. This derivation supplements the source answer.

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Common mistakes

Divide for ordering within each group and also for permuting equal-sized unlabelled groups. These are different sources of overcounting.

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