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Question Number 90531 by Mikael_786 last updated on 24/Apr/20
$${Please}\:{help}\:{me}\:{to}\:{find}\:{the}\:{value} \\ $$$${of}\:\:\:'{n}' \\ $$$${C}_{\mathrm{8}} ^{{n}+\mathrm{3}} ={C}_{\mathrm{20}} ^{{n}+\mathrm{3}} \\ $$
Commented by mr W last updated on 24/Apr/20
$${n}+\mathrm{3}=\mathrm{8}+\mathrm{20} \\ $$$$\Rightarrow{n}=\mathrm{25} \\ $$
Commented by jagoll last updated on 24/Apr/20
$$\frac{\left({n}+\mathrm{3}\right)!}{\mathrm{8}!\:\left({n}−\mathrm{5}\right)!}\:=\:\frac{\left({n}+\mathrm{3}\right)!}{\mathrm{20}!\:\left({n}−\mathrm{17}\right)!} \\ $$$$\mathrm{8}!\:\left({n}−\mathrm{5}\right)!\:=\:\mathrm{20}!\:\left({n}−\mathrm{17}\right)! \\ $$$${n}\:=\:\mathrm{25}\: \\ $$