How many ways are there to divide 11 people into 3 groups of 3 and one group of 2?
The right answer is 15400 but I can't get it
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$$\binom{11}{3}\cdot \binom{8}{3} \cdot \binom{5}{3} \over 3!$$ We divide by $3!$ because we dont care about order in group
Zereges
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Vladimir Fomenko
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You can use the \binom $\binom{n}{p}$ instead of the old notation $C_n^p$. – Xoff Jan 04 '15 at 09:56
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Imagine how many ways can you arrange 11 people in a row?
That is $11!$ Then let the first 3 be group 1, the next 3 be group 2, the next 3 be group 3 and the last 2 be group 4.
There are $3!$ ways to arrange each group of 3 and $2!$ ways to arrange the group of 2. As well as $3!$ ways to arrange the 3 groups of 3. So... $$\frac{11!}{3!\times3!\times3!\times2!\times3!}$$
EDIT: Volodymyr Fomenko's answer is equivalent and makes more sense than mine. $$\frac{11\times10\times9}{3!}\frac{8\times7\times6}{3!}\frac{5\times4\times3}{3!}\frac{2\times1}{2!}\frac{1}{3!}$$
turkeyhundt
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