I can't understand how: $$ \frac {2\times{^nC_2}}{5} $$
Equals:
$$ 2\times \frac {^nC_2}{5} $$
If we forget the combination and replace it with a $10$, the result is clearly different. $1$ in the first example and and $0.5$ in the second.
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2$\dfrac{2\times10}{5}=4$ and $2\times\dfrac{10}5=4$. It is not clear how you got $1$ or $0.5$. ${}\qquad{}$ – Michael Hardy Dec 24 '14 at 18:24
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What does the notation ${}^n C_2$ mean? Is that the binomial coefficient $\binom{n}{2}$ ("$n$ choose 2")? – Daniel Hast Dec 24 '14 at 18:24
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Forget that Michael, silly mistake from me. – Jet_Set_Willy Dec 24 '14 at 18:53
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Daniel, in Portugal we learn to write combinations that way. I didn't know it could also be represented that way. – Jet_Set_Willy Dec 24 '14 at 18:54
3 Answers
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Remember some fact about fractions: $\dfrac{a}{b}\times \dfrac{c}{d} = \dfrac{ac}{bd} \Rightarrow \dfrac{2\times \binom{n}{2}}{5} = \dfrac{2\times \binom{n}{2}}{1\times 5} = \dfrac{2}{1}\times \dfrac{\binom{n}{2}}{5} = 2\times \dfrac{\binom{n}{2}}{5}$
DeepSea
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$$\frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b \times d}$$ So, taking your example $$\frac{2\times10}{5\times 1}=\dfrac{2}{5}\times\dfrac{10}{1}=\dfrac{2}{1}\times\frac{10}{5}=\dfrac{10}{5}\times 2=\dfrac{2}{5}\times 10$$ As multiplication is commutative.
JEET TRIVEDI
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Using your example, $$ \frac{2\times 10}{5}=\frac{20}{5}=4$$ And $$ 2\times\frac{10}{5}=2\times 2=4$$ In general for any $c\not =0$, we have $$ \frac{a\times b}{c}=a\times\frac{b}{c}= b\times\frac{a}{c} $$
k170
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