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Question Number 171012 by mr W last updated on 06/Jun/22

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{five}\mathrm{digits}\mathrm{can}\mathrm{be}\mathrm{made}$$$$\mathrm{with}\mathrm{the}\mathrm{digits}1,2,3\mathrm{each}\mathrm{of}\mathrm{which}\mathrm{can}$$$$\mathrm{be}\mathrm{used}\mathrm{atmost}\mathrm{thrice}\mathrm{in}\mathrm{a}\mathrm{number}\mathrm{is}$$

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

$$a=numberofdigits1inthenumber$$$$b=numberofdigits2inthenumber$$$$c=numberofdigits3inthenumber$$$$0⩽a,b,c⩽3$$$$a+b+c=5$$$$coefficientof{x}^{5}in5!{(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!})}^3$$$$is210.\Rightarrow 210numbers$$

Commented by Tawa11 last updated on 06/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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