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There are 9 distinct chairs. How many ways are there to group these chairs into 3 groups of 3?

YEET
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1 Answers1

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Are the groups labelled?

The ways of forming a sequences with the chairs is $9!$. This doesn't change if we make it in a different way: let's say we first separate them into three labelled groups ($N$ ways, the number we want), then we arrange the order in each group ($(3!)^3$). Thus $$9!=N(3!)^3$$ $$N=\frac{9!}{(3!)^3}$$ If the groups are not labelled we can first label them and then repeat the process, obtaining $$N=\frac{9!}{(3!)^4}$$

Another way: how many ways to create 3 gruops of chair, each one with a number from 1 to 3? We can first divide the chairs into 3 groups of 3 (there are $N$ possibilities, the number we want to compute) and the we assign a number to each gruop ($3!$ ways).We can also choose 3 chairs for the first group, then we choose from the 6 remaining 3 for the second and choose 3 from the 3 left for the last group, thus we have $$3!N=\binom{9}{3}\binom{6}{3}\binom{3}{3}=\frac{9!}{(3!)^3}$$ $$N=\frac{9!}{(3!)^4}$$

Arararararagi-kun
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  • Confused on the labelling part but do understand the ‘choose’ part. Answer actually turned out to be N/3! And the solution told me becase ‘the number of chairs are the same and the order of the groups don’t matter’ which I’m confused about. But I’ll give your solution an upvote – YEET Oct 11 '19 at 21:14
  • N with ‘choose’ – YEET Oct 11 '19 at 21:22
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    @003264_ I see, so you are not considering the order of forming the groups. When you choose one first then the others two you are labelling them in the order they are choose, so you have to divide by 3!, which is the number of ways to order the 3 groups – Arararararagi-kun Oct 11 '19 at 21:24
  • I have edited the answer in order to make it clearer – Arararararagi-kun Oct 11 '19 at 21:32