Question Number 100444 by mr W last updated on 26/Jun/20

Find the number of five−digit numbers containing exactly three different digits? Examples: 12312, 12224

Answered by Rasheed.Sindhi last updated on 28/Jun/20

^• Let a,b,c are digits different from eachother. ^• First digit (from left): a ^• First two digits: a→aa ab (2nd digit may be same or different) ^• First three digits: aa→aaa aab ab→aba abb abc ^• First four digits: aaa→aaab aab→aaba aabb aabc aba→abaa abab abac abb→abba abbb abbc abc→abca abcb abcc ^• Five digits(all possible cases) aaab→aaabc aaba→aabac aabb→aabbc aabc→aabca aabcb aabcc abaa→abaac abab→ababc abac→abaca abacb abacc abba→abbac abbb→abbbc abbc→abbca abbcb abbcc abca→abcaa abcab abcac abcb→abcba abcbb abcbc abcc→abcca abccb abccc Total possible cases 25 ★In each case: (let U={0,1,...,9}) ^• first a can be filled in (U−{0}) 9 ways ^• first b (from left) can be filled in (U−{a}) 9 ways ^• first c (from left) can be filled in (U−{a,b}) 8 ways ^• each of other a′s,b′s,c′s can be filled in 1 way Total ways in each case 9×9×8=648 Total of 25 cases: 648×25=16200

Commented bymr W last updated on 04/Jul/20

thanks for solving sir!

Answered by mr W last updated on 28/Jun/20

i found following general method. let′s consider the general case: n−digit numbers with exactly m different digits (m≤10≤n). at first let′s treat digit 0 like the other digits, i.e. a number may also begin with 0. to select m digits from the 10 available digits (0,1,2,...,9) there are C_m ^(10) ways. let′s treat the n digit positions in a n−digit number as n boxes and we are going to fill these boxes with balls in m different colours and each box should obtain at least one ball. to do this there are m!{_m ^n } ways. {_m ^n } are the so−called stirling numbers of the second kind. {_m ^n } is the number of ways to partition a set of n objects into m non−empty subsets. for the m colours to represent the m different digits there are m! ways. therefore number of n−digit numbers which can be formed by exactly m different digits is C_m ^(10) m!{_m ^n }. but among these numbers there are some ones which begin with 0. we know there are the same many numbers begining with 0 as with 1 or with 2 etc. so the number of numbers which don′t begin with 0 is (9/(10))×C_m ^(10) m!{_m ^n }. this is the answer of our question. with n=5, m=3 we get (9/(10))×C_3 ^(10) ×3!×{_3 ^5 } =(9/(10))×120×6×25=16200

Commented bymr W last updated on 28/Jun/20

Commented bymr W last updated on 28/Jun/20

Commented byRasheed.Sindhi last updated on 28/Jun/20

Deep thinking & Nice research!

Answered by ajfour last updated on 29/Jun/20

Answer is< ^(10) C_3 (5!)=((10×9×8)/(1×2×3))×120 ⇒ answer < 14,400 (i think)

Commented bymr W last updated on 29/Jun/20

this is not correct sir. with 5 different digits you can form 12345 ⇒5!=120 numvers. but with three different digits 1,2,3 you can form more 5 digit numbers: 11123: ⇒((5!)/(3!)) 12223: ⇒((5!)/(3!)) 12333: ⇒((5!)/(3!)) 11223: ⇒((5!)/(2!2!)) 11233: ⇒((5!)/(2!2!)) 12233: ⇒((5!)/(2!2!)) ⇒3×((5!)/(3!))+3×((5!)/(2!2!))=150 [=3!{_3 ^5 }=6×25] > 120

Commented byajfour last updated on 02/Jul/20

thanks for the light sir, i shall try to arrive at the answer, my way too.

Answered by mr W last updated on 04/Jul/20

Classical way i also used: to select 3 digits from 10 there are C_3 ^(10) =120 ways. to build a five digit number with 3 different digits, say 1,2,3, there are following cases: case 1: one digit triple, two digits once each: 11123,22213,33312 ⇒3×((5!)/(3!)) ways case 2: one digit once, two digits twice each: 12233,21133,31122 ⇒3×((5!)/(2!2!)) ways ⇒totally C_3 ^(10) ×3×(((5!)/(3!))+((5!)/(2!2!)))=18000 but this includes also those numbers beginning with 0. to build a number beginning with 0, case 1: 01222,02221 ⇒2×((4!)/(3!))×C_2 ^9 case 2: 01122 ⇒((4!)/(2!2!))×C_2 ^9 case 3: 00112,00122 ⇒2×((4!)/(2!))×C_2 ^9 case 4: 00012 ⇒((4!)/(2!))×C_2 ^9 ⇒totally (2×((4!)/(3!))+((4!)/(2!2!))+2×((4!)/(2!))+((4!)/(2!)))×C_2 ^9 =1800 ⇒total valid five−digit numbers: 18000−1800=16200

Commented byRasheed.Sindhi last updated on 04/Jul/20

Nice approach:Easy to understand!