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The-number-of-five-digits-can-be-made-with-the-digits-1-2-3-each-of-which-can-be-used-atmost-thrice-in-a-number-is-




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

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