Question Number 173535 by mnjuly1970 last updated on 13/Jul/22
Answered by mr W last updated on 13/Jul/22
$${in}\:{expansion}\:{of}\:\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +...+{x}_{{n}} \right)^{{m}} \\ $$$${the}\:{general}\:{term}\:{is}\:{x}_{\mathrm{1}} ^{{k}_{\mathrm{1}} } {x}_{\mathrm{2}} ^{{k}_{\mathrm{2}} } {x}_{\mathrm{3}} ^{{k}_{\mathrm{3}} } ...{x}_{{n}} ^{{k}_{{n}} } \\ $$$${with}\:\mathrm{0}\leqslant{k}_{{i}} \leqslant{m}\:{and}\:{k}_{\mathrm{1}} +{k}_{\mathrm{2}} +{k}_{\mathrm{3}} +...+{k}_{{n}} ={m} \\ $$$${there}\:{are}\:{C}_{{m}} ^{{n}−\mathrm{1}+{m}} \:{possibilities}.\:{i}.{e}. \\ $$$${the}\:{number}\:{of}\:{distinct}\:{terms}\:{is} \\ $$$${C}_{{m}} ^{{n}−\mathrm{1}+{m}} .\: \\ $$$${in}\:{current}\:{case}:\:{n}=\mathrm{4},\:{m}=\mathrm{10} \\ $$$${number}\:{of}\:{distinct}\:{terms}\:{is}\:{C}_{\mathrm{10}} ^{\mathrm{13}} =\mathrm{286}. \\ $$
Commented by mnjuly1970 last updated on 13/Jul/22
$$\:\:\:\:{thanks}\:{alot}\:\:{Master}\:{W} \\ $$
Commented by Tawa11 last updated on 13/Jul/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$