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Question Number 161888 by henderson last updated on 23/Dec/21

$$\mathrm{help}\mathrm{me}!$$$$\mathrm{solve}\mathrm{this}\mathrm{one}:{\mathrm{C}}_{40}^{2\mathrm{n}}={\mathrm{C}}_{40}^{16+\mathrm{n}}$$

Commented by mr W last updated on 23/Dec/21

$$C_r^k=C_r^{r-k}$$$$2n=40-(16+n)$$$$\Rightarrow n=8$$

Commented by greg_ed last updated on 23/Dec/21

$$\mathrm{thank}\mathrm{you},\mathrm{sir}\mathrm{W}!$$

Commented by kapoorshah last updated on 24/Dec/21

$$wrong!!!$$$$k⩾r$$$$$$

Commented by mr W last updated on 24/Dec/21

$${don}^{\prime }tsaywrong!{it}^{\prime }saquestionof$$$$definition.notallcountriesuse$$$$thesamedefinition.insomecountries$$$$likefrance,russia,polen,chinaetc.$$$$peoplewrite{C}_n^{k}insteadof{C}_k^{n}for$$$$thesamething\frac{n!}{k!(n-k)!}.$$

Commented by Ar Brandon last updated on 24/Dec/21

$$\mathrm{Mr}\mathrm{W}\mathrm{is}\mathrm{right}$$$${\mathrm{C}}_n^k=\stackrel{n}{}{\mathrm{C}}_k=\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!(n-k)!}$$

Answered by Ar Brandon last updated on 23/Dec/21

$$2n=16+n\Rightarrow n=16$$$$C_{40}^{2\left(16\right)}={\mathrm{C}}_{40}^{16+16}$$

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