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


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

$F i n d t h e n u m b e r o f f i v e - d i g i t n u m b e r s$ $c o n t a i n i n g e x a c t l y t h r e e d i f f e r e n t$ $d i g i t s ? E x a m p l e s : 12312, 12224$
	Answered by Rasheed.Sindhi last updated on 28/Jun/20
	$^• Let a,b,c are digits different from eachother. ^• First digit (from left): a ^• First two digits: a→aa ab (2nd digit may be same or different) ^• First three digits: aa→aaa aab ab→aba abb abc ^• First four digits: aaa→aaab aab→aaba aabb aabc aba→abaa abab abac abb→abba abbb abbc abc→abca abcb abcc ^• Five digits(all possible cases) aaab→aaabc aaba→aabac aabb→aabbc aabc→aabca aabcb aabcc abaa→abaac abab→ababc abac→abaca abacb abacc abba→abbac abbb→abbbc abbc→abbca abbcb abbcc abca→abcaa abcab abcac abcb→abcba abcbb abcbc abcc→abcca abccb abccc Total possible cases 25 ★In each case: (let U={0,1,...,9}) ^• first a can be filled in (U−{0}) 9 ways ^• first b (from left) can be filled in (U−{a}) 9 ways ^• first c (from left) can be filled in (U−{a,b}) 8 ways ^• each of other a′s,b′s,c′s can be filled in 1 way Total ways in each case 9×9×8=648 Total of 25 cases: 648×25=16200$
	$^{∙} L e t a, b, c a r e d i g i t s d i f f e r e n t$ $f r o m e a c h o t h e r .$ $^{∙} F i r s t d i g i t (f r o m l e f t) : a$ $^{∙} F i r s t t w o d i g i t s : a \to a a a b$ $(2 n d d i g i t m a y b e s a m e o r d i f f e r e n t)$ $^{∙} F i r s t t h r e e d i g i t s :$ $a a \to a a a a a b$ $a b \to a b a a b b a b c$ $^{∙} F i r s t f o u r d i g i t s :$ $a a a \to a a a b$ $a a b \to a a b a a a b b a a b c$ $a b a \to a b a a a b a b a b a c$ $a b b \to a b b a a b b b a b b c$ $a b c \to a b c a a b c b a b c c$ $^{∙} F i v e d i g i t s (a l l p o s s i b l e c a s e s)$ $a a a b \to a a a b c$ $a a b a \to a a b a c$ $a a b b \to a a b b c$ $a a b c \to a a b c a a a b c b a a b c c$ $a b a a \to a b a a c$ $a b a b \to a b a b c$ $a b a c \to a b a c a a b a c b a b a c c$ $a b b a \to a b b a c$ $a b b b \to a b b b c$ $a b b c \to a b b c a a b b c b a b b c c$ $a b c a \to a b c a a a b c a b a b c a c$ $a b c b \to a b c b a a b c b b a b c b c$ $a b c c \to a b c c a a b c c b a b c c c$ $T o t a l p o s s i b l e c a s e s 25$ $★ I n e a c h c a s e :$ $(l e t U = {0, 1, . . ., 9})$ $^{∙} f i r s t a c a n b e f i l l e d i n$ $(U - {0}) 9 w a y s$ $^{∙} f i r s t b (f r o m l e f t) c a n b e f i l l e d$ $i n (U - {a}) 9 w a y s$ $^{∙} f i r s t c (f r o m l e f t) c a n b e f i l l e d$ $i n (U - {a, b}) 8 w a y s$ $^{∙} e a c h o f o t h e r a^{'} s, b^{'} s, c^{'} s c a n$ $b e f i l l e d i n 1 w a y$ $T o t a l w a y s i n e a c h c a s e$ $9 \times 9 \times 8 = 648$ $T o t a l o f 25 c a s e s :$ $648 \times 25 = 16200$
	Commented by mr W last updated on 04/Jul/20

	$t h a n k s f o r s o l v i n g s i r!$
	Answered by mr W last updated on 28/Jun/20
	$i found following general method. let′s consider the general case: n−digit numbers with exactly m different digits (m≤10≤n). at first let′s treat digit 0 like the other digits, i.e. a number may also begin with 0. to select m digits from the 10 available digits (0,1,2,...,9) there are C_m ^(10) ways. let′s treat the n digit positions in a n−digit number as n boxes and we are going to fill these boxes with balls in m different colours and each box should obtain at least one ball. to do this there are m!{_m ^n } ways. {_m ^n } are the so−called stirling numbers of the second kind. {_m ^n } is the number of ways to partition a set of n objects into m non−empty subsets. for the m colours to represent the m different digits there are m! ways. therefore number of n−digit numbers which can be formed by exactly m different digits is C_m ^(10) m!{_m ^n }. but among these numbers there are some ones which begin with 0. we know there are the same many numbers begining with 0 as with 1 or with 2 etc. so the number of numbers which don′t begin with 0 is (9/(10))×C_m ^(10) m!{_m ^n }. this is the answer of our question. with n=5, m=3 we get (9/(10))×C_3 ^(10) ×3!×{_3 ^5 } =(9/(10))×120×6×25=16200$
	$i f o u n d f o l l o w i n g g e n e r a l m e t h o d .$ ${l e t}^{'} s c o n s i d e r t h e g e n e r a l c a s e :$ $n - d i g i t n u m b e r s w i t h e x a c t l y m$ $d i f f e r e n t d i g i t s (m ⩽ 10 ⩽ n) .$ $a t f i r s t {l e t}^{'} s t r e a t d i g i t 0 l i k e t h e o t h e r$ $d i g i t s, i . e . a n u m b e r m a y a l s o b e g i n$ $w i t h 0 .$ $t o s e l e c t m d i g i t s f r o m t h e 10$ $a v a i l a b l e d i g i t s (0, 1, 2, . . ., 9) t h e r e a r e$ $C_{m}^{10} w a y s .$ ${l e t}^{'} s t r e a t t h e n d i g i t p o s i t i o n s i n a$ $n - d i g i t n u m b e r a s n b o x e s a n d w e$ $a r e g o i n g t o f i l l t h e s e b o x e s w i t h b a l l s$ $i n m d i f f e r e n t c o l o u r s a n d e a c h b o x$ $s h o u l d o b t a i n a t l e a s t o n e b a l l . t o d o$ $t h i s t h e r e a r e m! {_{m}^{n}} w a y s . {_{m}^{n}} a r e t h e$ $s o - c a l l e d s t i r l i n g n u m b e r s o f t h e$ $s e c o n d k i n d . {_{m}^{n}} i s t h e n u m b e r o f w a y s$ $t o p a r t i t i o n a s e t o f n o b j e c t s i n t o m$ $n o n - e m p t y s u b s e t s . f o r t h e m$ $c o l o u r s t o r e p r e s e n t t h e m d i f f e r e n t$ $d i g i t s t h e r e a r e m! w a y s . t h e r e f o r e$ $n u m b e r o f n - d i g i t n u m b e r s w h i c h$ $c a n b e f o r m e d b y e x a c t l y m d i f f e r e n t$ $d i g i t s i s C_{m}^{10} m! {_{m}^{n}} . b u t a m o n g t h e s e$ $n u m b e r s t h e r e a r e s o m e o n e s w h i c h$ $b e g i n w i t h 0 . w e k n o w t h e r e a r e t h e$ $s a m e m a n y n u m b e r s b e g i n i n g w i t h 0$ $a s w i t h 1 o r w i t h 2 e t c . s o t h e n u m b e r$ $o f n u m b e r s w h i c h {d o n}^{'} t b e g i n w i t h 0$ $i s \frac{9}{10} \times C_{m}^{10} m! {_{m}^{n}} . t h i s i s t h e a n s w e r$ $o f o u r q u e s t i o n .$ $w i t h n = 5, m = 3 w e g e t$ $\frac{9}{10} \times C_{3}^{10} \times 3! \times {_{3}^{5}}$ $= \frac{9}{10} \times 120 \times 6 \times 25 = 16200$
	Commented by mr W last updated on 28/Jun/20

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

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

	$D e e p t h i n k i n g & N i c e r e s e a r c h!$
	Answered by ajfour last updated on 29/Jun/20

	$A n s w e r i s <^{10} C_{3} (5!) = \frac{10 \times 9 \times 8}{1 \times 2 \times 3} \times 120$ $\Rightarrow a n s w e r < 14, 400 (i t h i n k)$
	Commented by mr W last updated on 29/Jun/20
	$this is not correct sir. with 5 different digits you can form 12345 ⇒5!=120 numvers. but with three different digits 1,2,3 you can form more 5 digit numbers: 11123: ⇒((5!)/(3!)) 12223: ⇒((5!)/(3!)) 12333: ⇒((5!)/(3!)) 11223: ⇒((5!)/(2!2!)) 11233: ⇒((5!)/(2!2!)) 12233: ⇒((5!)/(2!2!)) ⇒3×((5!)/(3!))+3×((5!)/(2!2!))=150 [=3!{_3 ^5 }=6×25] > 120$
	$t h i s i s n o t c o r r e c t s i r .$ $w i t h 5 d i f f e r e n t d i g i t s y o u c a n f o r m$ $12345 \Rightarrow 5! = 120 n u m v e r s .$ $b u t w i t h t h r e e d i f f e r e n t d i g i t s 1, 2, 3$ $y o u c a n f o r m m o r e 5 d i g i t n u m b e r s :$ $11123 : \Rightarrow \frac{5!}{3!}$ $12223 : \Rightarrow \frac{5!}{3!}$ $12333 : \Rightarrow \frac{5!}{3!}$ $11223 : \Rightarrow \frac{5!}{2! 2!}$ $11233 : \Rightarrow \frac{5!}{2! 2!}$ $12233 : \Rightarrow \frac{5!}{2! 2!}$ $\Rightarrow 3 \times \frac{5!}{3!} + 3 \times \frac{5!}{2! 2!} = 150 [= 3! {_{3}^{5}} = 6 \times 25] > 120$
	Commented by ajfour last updated on 02/Jul/20

	$t h a n k s f o r t h e l i g h t s i r, i s h a l l t r y$ $t o a r r i v e a t t h e a n s w e r, m y w a y t o o .$
	Answered by mr W last updated on 04/Jul/20

	$C l a s s i c a l w a y i a l s o u s e d :$ $t o s e l e c t 3 d i g i t s f r o m 10 t h e r e a r e$ $C_{3}^{10} = 120 w a y s .$ $t o b u i l d a f i v e d i g i t n u m b e r w i t h 3$ $d i f f e r e n t d i g i t s, s a y 1, 2, 3, t h e r e a r e$ $f o l l o w i n g c a s e s :$ $c a s e 1 : o n e d i g i t t r i p l e, t w o d i g i t s$ $o n c e e a c h :$ $11123, 22213, 33312 \Rightarrow 3 \times \frac{5!}{3!} w a y s$ $c a s e 2 : o n e d i g i t o n c e, t w o d i g i t s$ $t w i c e e a c h :$ $12233, 21133, 31122 \Rightarrow 3 \times \frac{5!}{2! 2!} w a y s$ $\Rightarrow t o t a l l y C_{3}^{10} \times 3 \times (\frac{5!}{3!} + \frac{5!}{2! 2!}) = 18000$ $b u t t h i s i n c l u d e s a l s o t h o s e n u m b e r s$ $b e g i n n i n g w i t h 0 .$ $t o b u i l d a n u m b e r b e g i n n i n g w i t h 0,$ $c a s e 1 : 01222, 02221 \Rightarrow 2 \times \frac{4!}{3!} \times C_{2}^{9}$ $c a s e 2 : 01122 \Rightarrow \frac{4!}{2! 2!} \times C_{2}^{9}$ $c a s e 3 : 00112, 00122 \Rightarrow 2 \times \frac{4!}{2!} \times C_{2}^{9}$ $c a s e 4 : 00012 \Rightarrow \frac{4!}{2!} \times C_{2}^{9}$ $\Rightarrow t o t a l l y (2 \times \frac{4!}{3!} + \frac{4!}{2! 2!} + 2 \times \frac{4!}{2!} + \frac{4!}{2!}) \times C_{2}^{9} = 1800$ $\Rightarrow t o t a l v a l i d f i v e - d i g i t n u m b e r s :$ $18000 - 1800 = 16200$
	Commented by Rasheed.Sindhi last updated on 04/Jul/20

	$N i c e a p p r o a c h : E a s y t o u n d e r s t a n d!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum