Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

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.

$$\left({a}\right)\:\mathrm{Strictly}\:\mathrm{increasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{12568}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{only}\:{one}\:{time}.\:\mathrm{0}\:{can}\:{not}\:{be} \\ $$$${used}.\:{so}\:{we}\:{can}\:{only}\:{select}\:\mathrm{5}\:{from} \\ $$$$\mathrm{9}\:{digits}. \\ $$$${C}_{\mathrm{5}} ^{\mathrm{9}} =\mathrm{126}. \\ $$$$ \\ $$$$\left({b}\right)\:\mathrm{Strictly}\:\mathrm{decreasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{96540}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{only}\:{one}\:{time}.\:{we}\:{can}\:{select}\: \\ $$$$\mathrm{5}\:{from}\:\mathrm{10}\:{digits}. \\ $$$${C}_{\mathrm{5}} ^{\mathrm{10}} =\mathrm{252}. \\ $$$$ \\ $$$$\left({c}\right)\:\mathrm{Increasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{12248}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{more}\:{times}.\:\mathrm{0}\:{can}\:{not}\:{be}\:{used}. \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...\right)^{\mathrm{9}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{9}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{8}} ^{{k}+\mathrm{8}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{is} \\ $$$${C}_{\mathrm{8}} ^{\mathrm{5}+\mathrm{8}} =\mathrm{1287}. \\ $$$$ \\ $$$$\left({d}\right)\:{D}\mathrm{ecreasing}\:\mathrm{digits} \\ $$$${numbers}\:{like}\:\mathrm{85521}.\:{each}\:{digit}\:{may} \\ $$$${occur}\:{more}\:{times}. \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...\right)^{\mathrm{10}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{10}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{9}} ^{{k}+\mathrm{9}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{is} \\ $$$${C}_{\mathrm{9}} ^{\mathrm{5}+\mathrm{9}} =\mathrm{2002}. \\ $$$${since}\:\mathrm{00000}\:{is}\:{not}\:{a}\:{valid}\:{number}, \\ $$$${so}\:{the}\:{answer}\:{is}\:\mathrm{2002}−\mathrm{1}=\mathrm{2001}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com