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Question Number 209794 by Ismoiljon_008 last updated on 21/Jul/24

        13456622577532674 how many 5-digit     numbers can be made from these numbers?     help please

13456622577532674howmany5digitnumberscanbemadefromthesenumbers?helpplease

Answered by MM42 last updated on 21/Jul/24

1→1  /  2→3  /  3→2 /  4→2/ 5→3 /6→3 / 7→3   abcde⇒7×6×5×4×3=2520  aabcd⇒ ((6),(1) )× ((6),(3) )×((5!)/(2!))=7200  aabbc⇒ ((6),(2) )× ((5),(1) )×((5!)/(2!2!))=2250  aaabc⇒ ((4),(1) )× ((6),(2) )×((5!)/(3!))=1200  aaabb⇒ ((4),(1) )× ((5),(1) )×((5!)/(3!2!))=200  ans=13370 ✓

11/23/32/42/53/63/73abcde7×6×5×4×3=2520aabcd(61)×(63)×5!2!=7200aabbc(62)×(51)×5!2!2!=2250aaabc(41)×(62)×5!3!=1200aaabb(41)×(51)×5!3!2!=200ans=13370

Commented by Ismoiljon_008 last updated on 22/Jul/24

   thank you very much

thankyouverymuch

Commented by mr W last updated on 22/Jul/24

method is right, but answer is wrong.  abcde⇒7×6×5×4×3=2520 ✓  aabcd⇒ ((5),(1) )× ((6),(3) )×((5!)/(2!))=6000  aabbc⇒ ((5),(2) )× ((5),(1) )×((5!)/(2!2!))=1500  aaabc⇒ ((4),(1) )× ((6),(2) )×((5!)/(3!))=1200 ✓  aaabb⇒ ((4),(1) )× ((4),(1) )×((5!)/(3!2!))=160  totally: 11380

methodisright,butansweriswrong.abcde7×6×5×4×3=2520aabcd(51)×(63)×5!2!=6000aabbc(52)×(51)×5!2!2!=1500aaabc(41)×(62)×5!3!=1200aaabb(41)×(41)×5!3!2!=160totally:11380

Answered by mr W last updated on 22/Jul/24

an other path using generating function:  1, 3  44  222, 555, 666, 777  the answer is the coef. of term x^5  of  the expansion  5!(1+x)^2 (1+x+(x^2 /(2!)))(1+x+(x^2 /(2!))+(x^3 /(3!)))^4   which is 11380.

anotherpathusinggeneratingfunction:1,344222,555,666,777theansweristhecoef.oftermx5oftheexpansion5!(1+x)2(1+x+x22!)(1+x+x22!+x33!)4whichis11380.

Commented by mr W last updated on 22/Jul/24

Commented by mr W last updated on 22/Jul/24

if the question is “how many 7  digit numbers can be made”, then  the answer is the coef. of term x^7  of  7!(1+x)^2 (1+x+(x^2 /(2!)))(1+x+(x^2 /(2!))+(x^3 /(3!)))^4   which is 358680.

ifthequestionishowmany7digitnumberscanbemade,thentheansweristhecoef.oftermx7of7!(1+x)2(1+x+x22!)(1+x+x22!+x33!)4whichis358680.

Commented by mr W last updated on 22/Jul/24

Commented by mr W last updated on 22/Jul/24

or we just consider the generating  function  (1+x)^2 (1+x+(x^2 /(2!)))(1+x+(x^2 /(2!))+(x^3 /(3!)))^4   then the number of k−digit numbers  is k!×coefficient of term x^k .

orwejustconsiderthegeneratingfunction(1+x)2(1+x+x22!)(1+x+x22!+x33!)4thenthenumberofkdigitnumbersisk!×coefficientoftermxk.

Commented by mr W last updated on 22/Jul/24

Commented by mr W last updated on 22/Jul/24

examples:  4 digit numbers: 4!×79=1896  5 digit numbers: 5!×((569)/6)=11380  6 digit numbers: 6!×((1091)/(12))=65460  7 digit numbers: 7!×((427)/6)=358680  10 digit numbers: 10!×((9577)/(864))=40223400  etc.

examples:4digitnumbers:4!×79=18965digitnumbers:5!×5696=113806digitnumbers:6!×109112=654607digitnumbers:7!×4276=35868010digitnumbers:10!×9577864=40223400etc.

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