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In how many ways can $15$ different things be divided into three groups containing $7,5$ and $3$ things.

According to me it should be $3! \times ^{15}C_7 \times ^8C_5\times ^3C_3 =2162160 $ (note: $3!$ occurs to include the possibility of different groups (say GroupA , Group B , Group C) getting any particular combination ($7,5$ or $3$))

But answer given is $360360$.

Please tell me the fault in my reasoning.

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It should be the multinomial coefficient $$\frac{15!}{7!5!3!}$$ which indeed is $360360$. You have inflated the total by a factor $3!$ by adding labels to the three subsets (presumably not asked for in the question).

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    $\begingroup$ But that is a possibility. WHy not include that? $\endgroup$ – Archer Dec 18 '17 at 7:16
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    $\begingroup$ Are we supposed to assume that the groups are indistinguishable? $\endgroup$ – Archer Dec 18 '17 at 7:17
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    $\begingroup$ @Abcd Of course they are distinguishable; they have different sizes. $\endgroup$ – Lord Shark the Unknown Dec 18 '17 at 7:27
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    $\begingroup$ Then why no multiplication by $3!$? $\endgroup$ – Archer Dec 18 '17 at 7:30
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    $\begingroup$ @Abcd If you do $\binom{15}{7}\binom{8}{5}\binom{3}{3}$, it already takes into account all possible options. $\endgroup$ – samjoe Dec 18 '17 at 7:38

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