# Question-134255

Question Number 134255 by shaker last updated on 01/Mar/21
Commented by Dwaipayan Shikari last updated on 01/Mar/21
$${C}_{{m}} ^{{n}} \:\:\:\:\:\left({n}<{m}\right)\:{is}\:{not}\:{defined} \\$$
Commented by mr W last updated on 01/Mar/21
$${i}\:{think}\:{he}\:{means}\:{C}_{{n}} ^{{m}} \:\left({n}<{m}\right).\:{in}\:{some}\: \\$$$${countries}\:{people}\:{write}\:{C}_{{m}} ^{{n}} \:{instead}\:{of} \\$$$${C}_{{n}} ^{{m}} . \\$$
Answered by EDWIN88 last updated on 01/Mar/21
$$\mathrm{let}\::\:\mathrm{T}\:=\:\frac{\frac{\mathrm{100}×\mathrm{99}}{\mathrm{2}×\mathrm{1}}+\frac{\mathrm{1000}×\mathrm{999}}{\mathrm{2}×\mathrm{1}}}{\frac{\mathrm{1000}×\mathrm{999}}{\mathrm{2}×\mathrm{1}}+\frac{\mathrm{100}×\mathrm{99}}{\mathrm{2}×\mathrm{1}}}\:=\:\mathrm{1} \\$$$$\mathrm{let}\:\mathrm{Q}\:=\:\frac{\mathrm{1}+\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}}{\mathrm{3}×\mathrm{2}×\mathrm{1}}\:+\:\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}}{\mathrm{3}×\mathrm{2}×\mathrm{1}}−\frac{\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}}{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}}{\mathrm{1}+\frac{\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}+\frac{\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}−\frac{\mathrm{11}×\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}} \\$$
Answered by Ñï= last updated on 01/Mar/21
$${A}=\frac{{C}_{\mathrm{100}} ^{\mathrm{2}} +{C}_{\mathrm{1000}} ^{\mathrm{2}} }{{C}_{\mathrm{1000}} ^{\mathrm{2}} +{C}_{\mathrm{100}} ^{\mathrm{2}} }+\frac{\mathrm{1}+{C}_{\mathrm{8}} ^{\mathrm{4}} −{C}_{\mathrm{8}} ^{\mathrm{4}} }{\mathrm{1}+{C}_{\mathrm{11}} ^{\mathrm{6}} −{C}_{\mathrm{11}} ^{\mathrm{6}} } \\$$$$=\mathrm{1}+\mathrm{1} \\$$$$=\mathrm{2} \\$$$${C}_{{n}} ^{{k}} +{C}_{{n}} ^{{k}−\mathrm{1}} ={C}_{{n}+\mathrm{1}} ^{{k}} \Leftrightarrow\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix} \\$$