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Question Number 171012 by mr W last updated on 06/Jun/22
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{five}\:\mathrm{digits}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\mathrm{each}\:\mathrm{of}\:\mathrm{which}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{used}\:\mathrm{atmost}\:\mathrm{thrice}\:\mathrm{in}\:\mathrm{a}\:\mathrm{number}\:\mathrm{is} \\ $$
Commented by otchereabdullai@gmail.com last updated on 07/Jun/22
$$\mathrm{nice}\:\mathrm{question}\:+\:\mathrm{nice}\:\mathrm{solution}! \\ $$
Answered by mr W last updated on 06/Jun/22
$${a}={number}\:{of}\:{digits}\:\mathrm{1}\:{in}\:{the}\:{number} \\ $$$${b}={number}\:{of}\:{digits}\:\mathrm{2}\:{in}\:{the}\:{number} \\ $$$${c}={number}\:{of}\:{digits}\:\mathrm{3}\:{in}\:{the}\:{number} \\ $$$$\mathrm{0}\leqslant{a},{b},{c}\leqslant\mathrm{3} \\ $$$${a}+{b}+{c}=\mathrm{5} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{5}} \:{in}\:\mathrm{5}!\left(\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\right)^{\mathrm{3}} \\ $$$${is}\:\mathrm{210}.\:\Rightarrow\mathrm{210}\:{numbers} \\ $$
Commented by Tawa11 last updated on 06/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$