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 125548 by Don08q last updated on 12/Dec/20

How many four−digit numbers   between 2000 and 4000 can be formed  from the numbers 0, 1, 2, 3, 4 and 5   if repitition is allowed?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{four}−\mathrm{digit}\:\mathrm{numbers}\: \\ $$$$\mathrm{between}\:\mathrm{2000}\:\mathrm{and}\:\mathrm{4000}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\mathrm{and}\:\mathrm{5}\: \\ $$$$\mathrm{if}\:\mathrm{repitition}\:\mathrm{is}\:\mathrm{allowed}? \\ $$

Commented by liberty last updated on 13/Dec/20

let the number ABCD where 2000<ABCD<4000  case(1) A_(2,3  [) B_(1,2,3,4,5]) CD = 2×5×6^2  = 360  case(2) A_(2,3)  B_0  C_([1,2,3,4,5])  D = 2×1×5×6=60  case(3) A_(2,3)  B_0  C_0  D_([ 1,2,3,4,5 ]) = 2×1×1×5=10  case(4) A_3  B_0  C_0  D_0  = 1×1×1×1=1  totally = 431

$${let}\:{the}\:{number}\:{ABCD}\:{where}\:\mathrm{2000}<{ABCD}<\mathrm{4000} \\ $$$${case}\left(\mathrm{1}\right)\:\underset{\mathrm{2},\mathrm{3}\:\:\left[\right.} {{A}}\underset{\left.\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right]} {{B}CD}\:=\:\mathrm{2}×\mathrm{5}×\mathrm{6}^{\mathrm{2}} \:=\:\mathrm{360} \\ $$$${case}\left(\mathrm{2}\right)\:\underset{\mathrm{2},\mathrm{3}} {{A}}\:\underset{\mathrm{0}} {{B}}\:\underset{\left[\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right]} {{C}}\:{D}\:=\:\mathrm{2}×\mathrm{1}×\mathrm{5}×\mathrm{6}=\mathrm{60} \\ $$$${case}\left(\mathrm{3}\right)\:\underset{\mathrm{2},\mathrm{3}} {{A}}\:\underset{\mathrm{0}} {{B}}\:\underset{\mathrm{0}} {{C}}\:\underset{\left[\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\:\right]} {{D}}=\:\mathrm{2}×\mathrm{1}×\mathrm{1}×\mathrm{5}=\mathrm{10} \\ $$$${case}\left(\mathrm{4}\right)\:\underset{\mathrm{3}} {{A}}\:\underset{\mathrm{0}} {{B}}\:\underset{\mathrm{0}} {{C}}\:\underset{\mathrm{0}} {{D}}\:=\:\mathrm{1}×\mathrm{1}×\mathrm{1}×\mathrm{1}=\mathrm{1} \\ $$$${totally}\:=\:\mathrm{431}\: \\ $$

Commented by bemath last updated on 12/Dec/20

i think your answer correct sir

$${i}\:{think}\:{your}\:{answer}\:{correct}\:{sir} \\ $$

Answered by Olaf last updated on 12/Dec/20

The first digit is 2 or 3 ⇒ 2 possibilities.  For the 3 other digits ⇒ 6 possibilities.  2×6^3  = 432  Here we obtain the possibilities  between 2000 and 3555.  But 4000 is a possibility too.  ⇒ 432+1 = 433 possibilities.

$$\mathrm{The}\:\mathrm{first}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{2}\:\mathrm{or}\:\mathrm{3}\:\Rightarrow\:\mathrm{2}\:\mathrm{possibilities}. \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{3}\:\mathrm{other}\:\mathrm{digits}\:\Rightarrow\:\mathrm{6}\:\mathrm{possibilities}. \\ $$$$\mathrm{2}×\mathrm{6}^{\mathrm{3}} \:=\:\mathrm{432} \\ $$$$\mathrm{Here}\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{the}\:\mathrm{possibilities} \\ $$$$\mathrm{between}\:\mathrm{2000}\:\mathrm{and}\:\mathrm{3555}. \\ $$$$\mathrm{But}\:\mathrm{4000}\:\mathrm{is}\:\mathrm{a}\:\mathrm{possibility}\:\mathrm{too}. \\ $$$$\Rightarrow\:\mathrm{432}+\mathrm{1}\:=\:\mathrm{433}\:\mathrm{possibilities}. \\ $$

Commented by malwaan last updated on 12/Dec/20

mr olaf  I think the right answer is   432−2= 430  the question is between 2000  and 4000  If  the question  was from 2000   to 4000 then the answer will  be 432

$${mr}\:{olaf} \\ $$$${I}\:{think}\:{the}\:{right}\:{answer}\:{is}\: \\ $$$$\mathrm{432}−\mathrm{2}=\:\mathrm{430} \\ $$$${the}\:{question}\:{is}\:{between}\:\mathrm{2000} \\ $$$${and}\:\mathrm{4000} \\ $$$${If}\:\:{the}\:{question}\:\:{was}\:{from}\:\mathrm{2000}\: \\ $$$${to}\:\mathrm{4000}\:{then}\:{the}\:{answer}\:{will} \\ $$$${be}\:\mathrm{432} \\ $$

Commented by malwaan last updated on 12/Dec/20

sorry   Iam not sure

$${sorry}\: \\ $$$${Iam}\:{not}\:{sure} \\ $$

Commented by mr W last updated on 12/Dec/20

for 2000≤x≤4000 the answer is  2×6^3 +1=433  for 2000<x<4000 the answer is  2×6^3 −1=431

$${for}\:\mathrm{2000}\leqslant{x}\leqslant\mathrm{4000}\:{the}\:{answer}\:{is} \\ $$$$\mathrm{2}×\mathrm{6}^{\mathrm{3}} +\mathrm{1}=\mathrm{433} \\ $$$${for}\:\mathrm{2000}<{x}<\mathrm{4000}\:{the}\:{answer}\:{is} \\ $$$$\mathrm{2}×\mathrm{6}^{\mathrm{3}} −\mathrm{1}=\mathrm{431} \\ $$

Commented by Don08q last updated on 12/Dec/20

Yes. 431 is the correct answer. ✓✓

$$\mathrm{Yes}.\:\mathrm{431}\:\mathrm{is}\:\mathrm{the}\:\boldsymbol{\mathrm{correct}}\:\mathrm{answer}.\:\checkmark\checkmark \\ $$

Commented by Olaf last updated on 13/Dec/20

Sorry, my english is not fluent.  In my language, “between” means ≤

$$\mathrm{Sorry},\:\mathrm{my}\:\mathrm{english}\:\mathrm{is}\:\mathrm{not}\:\mathrm{fluent}. \\ $$$$\mathrm{In}\:\mathrm{my}\:\mathrm{language},\:``\mathrm{between}''\:\mathrm{means}\:\leqslant \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com