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Question Number 171011 by mr W last updated on 06/Jun/22
How many 5-digit numbers from the digits {0, 1, ....., 9} have? (i) Strictly increasing digits (ii) Strictly increasing or decreasing digits (iii) Increasing digits (iv) Increasing or decreasing digits
$$\mathrm{How}\:\mathrm{many}\:\mathrm{5}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{digits}\:\left\{\mathrm{0},\:\mathrm{1},\:…..,\:\mathrm{9}\right\}\:\mathrm{have}? \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Strictly}\:\mathrm{increasing}\:\mathrm{digits} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Strictly}\:\mathrm{increasing}\:\mathrm{or}\:\mathrm{decreasing} \\ $$$$\mathrm{digits} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{Increasing}\:\mathrm{digits} \\ $$$$\left(\mathrm{iv}\right)\:\mathrm{Increasing}\:\mathrm{or}\:\mathrm{decreasing}\:\mathrm{digits} \\ $$
Answered by aleks041103 last updated on 07/Jun/22
let: • the first digit be a • the second − a+b • the 3rd − a+b+c • the 4th − a+b+c+d • the 5th − a+b+c+d+e (i) ⇔ a+b+c+d+e≤9, a,...,e∈{1,...,9} let x_1 =a−1,...,x_5 =e−1 ⇒Σ_(i=1) ^5 x_i ≤4 where x_i ∈{0,...,8} let x_6 =4−Σ_(i=1) ^5 x_i ∈{0,...,4} ⇒we′ve reduced the problem to Σ_(i=1) ^6 x_i =4 this can be done in: C_4 ^(6+4−1) =C_4 ^9 =((9.8.7.6)/(1.2.3.4))=9.7.2=126 (ii) ... (iv) can ve done by analogy
$${let}: \\ $$$$\bullet\:{the}\:{first}\:{digit}\:{be}\:{a} \\ $$$$\bullet\:{the}\:{second}\:−\:{a}+{b} \\ $$$$\bullet\:{the}\:\mathrm{3}{rd}\:−\:{a}+{b}+{c} \\ $$$$\bullet\:{the}\:\mathrm{4}{th}\:−\:{a}+{b}+{c}+{d} \\ $$$$\bullet\:{the}\:\mathrm{5}{th}\:−\:{a}+{b}+{c}+{d}+{e} \\ $$$$\left({i}\right)\:\Leftrightarrow\:{a}+{b}+{c}+{d}+{e}\leqslant\mathrm{9},\:{a},…,{e}\in\left\{\mathrm{1},…,\mathrm{9}\right\} \\ $$$${let}\:{x}_{\mathrm{1}} ={a}−\mathrm{1},…,{x}_{\mathrm{5}} ={e}−\mathrm{1} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}{x}_{{i}} \leqslant\mathrm{4} \\ $$$${where}\:{x}_{{i}} \in\left\{\mathrm{0},…,\mathrm{8}\right\} \\ $$$${let}\:{x}_{\mathrm{6}} =\mathrm{4}−\underset{{i}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}{x}_{{i}} \in\left\{\mathrm{0},…,\mathrm{4}\right\} \\ $$$$\Rightarrow{we}'{ve}\:{reduced}\:{the}\:{problem}\:{to} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{x}_{{i}} =\mathrm{4} \\ $$$${this}\:{can}\:{be}\:{done}\:{in}:\:{C}_{\mathrm{4}} ^{\mathrm{6}+\mathrm{4}−\mathrm{1}} ={C}_{\mathrm{4}} ^{\mathrm{9}} =\frac{\mathrm{9}.\mathrm{8}.\mathrm{7}.\mathrm{6}}{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}}=\mathrm{9}.\mathrm{7}.\mathrm{2}=\mathrm{126} \\ $$$$\left({ii}\right)\:…\:\left({iv}\right)\:{can}\:{ve}\:{done}\:{by}\:{analogy} \\ $$$$ \\ $$
Commented by mr W last updated on 08/Jun/22
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 09/Jun/22
(a) Strictly increasing digits numbers like 12568. each digit may occur only one time. 0 can not be used. so we can only select 5 from 9 digits. C_5 ^9 =126. (b) Strictly decreasing digits numbers like 96540. each digit may occur only one time. we can select 5 from 10 digits. C_5 ^(10) =252. (c) Increasing digits numbers like 12248. each digit may occur more times. 0 can not be used. (1+x+x^2 +...)^9 =(1/((1−x)^9 ))=Σ_(k=0) ^∞ C_8 ^(k+8) x^k coef. of x^5 is C_8 ^(5+8) =1287. (d) Decreasing digits numbers like 85521. each digit may occur more times. (1+x+x^2 +...)^(10) =(1/((1−x)^(10) ))=Σ_(k=0) ^∞ C_9 ^(k+9) x^k coef. of x^5 is C_9 ^(5+9) =2002. since 00000 is not a valid number, so the answer is 2002−1=2001.
$$\left({a}\right)\:\mathrm{Strictly}\:\mathrm{increasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{12568}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{only}\:{one}\:{time}.\:\mathrm{0}\:{can}\:{not}\:{be} \\ $$$${used}.\:{so}\:{we}\:{can}\:{only}\:{select}\:\mathrm{5}\:{from} \\ $$$$\mathrm{9}\:{digits}. \\ $$$${C}_{\mathrm{5}} ^{\mathrm{9}} =\mathrm{126}. \\ $$$$ \\ $$$$\left({b}\right)\:\mathrm{Strictly}\:\mathrm{decreasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{96540}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{only}\:{one}\:{time}.\:{we}\:{can}\:{select}\: \\ $$$$\mathrm{5}\:{from}\:\mathrm{10}\:{digits}. \\ $$$${C}_{\mathrm{5}} ^{\mathrm{10}} =\mathrm{252}. \\ $$$$ \\ $$$$\left({c}\right)\:\mathrm{Increasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{12248}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{more}\:{times}.\:\mathrm{0}\:{can}\:{not}\:{be}\:{used}. \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{9}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{9}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{8}} ^{{k}+\mathrm{8}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{is} \\ $$$${C}_{\mathrm{8}} ^{\mathrm{5}+\mathrm{8}} =\mathrm{1287}. \\ $$$$ \\ $$$$\left({d}\right)\:{D}\mathrm{ecreasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{85521}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{more}\:{times}. \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{10}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{10}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{9}} ^{{k}+\mathrm{9}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{is} \\ $$$${C}_{\mathrm{9}} ^{\mathrm{5}+\mathrm{9}} =\mathrm{2002}. \\ $$$${since}\:\mathrm{00000}\:{is}\:{not}\:{a}\:{valid}\:{number}, \\ $$$${so}\:{the}\:{answer}\:{is}\:\mathrm{2002}−\mathrm{1}=\mathrm{2001}. \\ $$

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