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Question Number 41824 by Tawa1 last updated on 13/Aug/18

$$\mathrm{if}\stackrel{\mathrm{n}}{}{\mathrm{C}}_5=\stackrel{\mathrm{n}}{}{\mathrm{C}}_{11}\mathrm{find}\stackrel{18}{}{\mathrm{C}}_{\mathrm{n}}$$

Answered by $@ty@m last updated on 13/Aug/18

$${}^n{C}_r={}^n{C}_{n-r}$$$$\Rightarrow r=5$$$$\mathrm{\&}n-r=11\Rightarrow n=16$$$$\Rightarrow {}^{18}{C}_n={}^{18}{C}_{16}={}^{18}{C}_2=153$$

Commented by Tawa1 last updated on 13/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Aug/18

$$\frac{n\times (n-1)\times (n-3)\times (n-4)}{5!}=\frac{n\times (n-1)...\times (n-10)}{11!}$$$$(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)=11\times 10\times 9\times 8\times 7\times 6$$$$n=16$$$$18c_{16}=\frac{18\times 17}{2}=153$$

Commented by Tawa1 last updated on 13/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by malwaan last updated on 13/Aug/18

$$\mathrm{n}=5+11=16$$$$\therefore \underset{\mathrm{n}}{\stackrel{18}{\mathrm{C}}}=\underset{16}{\stackrel{18}{\mathrm{C}}}=\frac{18\times 17}{2\times 1}=9\times 17$$$$=153$$

Commented by Tawa1 last updated on 13/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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