How many 6 digit numbers exist whose digits have exactly the sum 13? for example 120505 is such a number.


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Question Number 102816 by mr W last updated on 11/Jul/20

$H o w m a n y 6 d i g i t n u m b e r s e x i s t$ $w h o s e d i g i t s h a v e e x a c t l y t h e s u m 13 ?$ $f o r e x a m p l e 120505 i s s u c h a n u m b e r .$
Commented by Rasheed.Sindhi last updated on 11/Jul/20

$Universal Set$ ${0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3,$ $4, 4, 4, 5, 5, 6, 6, 7, 8, 9}$
Commented by aurpeyz last updated on 11/Jul/20

$660100 508000 309000 . . . it will be much o . how should this be [done ?$
Commented by mr W last updated on 11/Jul/20

$R a s h e e d s i r : i {d o n}^{'} t k n o w m u c h$ $a b o u t s e t m e t h o d s . c a n y o u g i v e s o m e$ $e x p l a n a t i o n a n d h o w t o a r r i v e a t t h e$ $r e s u l t ? t h a n k s!$
Commented by mr W last updated on 11/Jul/20

$i s e e, t h a n k s! d o y o u h a v e o t h e r i d e a s ?$
Commented by Rasheed.Sindhi last updated on 11/Jul/20

$S i r, a c t u a l l y {i t}^{'} s n o t m u c h$ $h e l p f u l l, I t h i n k n o w . I o n l y$ $e x t e n d e d t h e s e t {0, 1, 2, . . ., 9} b y$ $w r i t i n g m a x i m u m p o s s i b l e$ $o c c u r i g s o f e a c h d i g i t i n t h e$ $r e q u i r e d n u m b e r s .$
Commented by mr W last updated on 11/Jul/20

$i g o t 6027 s u c h n u m b e r s .$
Commented by Rasheed.Sindhi last updated on 11/Jul/20

$S i r I t h o u g h t t o a t t a c k t h e$ $p r o b l e m a s f o l l o w s :$ $(i) T o d e t e r m i n e t h e n u m b e r$ $o f p a r t i o n s (i n a n o r d e r)$ $o f 13 u s i n g a b o v e^{'} u n i v e r s a l {s e t}^{'}$ ${0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3,$ $4, 4, 4, 5, 5, 6, 6, 7, 8, 9}$ $(i i) T o d e t e r m i n e n u m b e r o f$ $^{'} {p e r m u t a t i o n s}^{'} o f e a c h$ $p a r t i t i o n s .$ $(i i i) A f t e r t h a n n u m b e r o f$ $e x c l u d i n g u n w a n t e d n u m b e r s .$ $I a m n o t c o n f i d e n t o f t h e s e$ $i d e a s . Y o u c a n d o b e t t e r .$
Commented by prakash jain last updated on 11/Jul/20
This method i think is known as generating function method, used in problem involving coins etc.
Commented by prakash jain last updated on 11/Jul/20

$Thanks . I will recheck calculation$ $for both methods .$
Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

$Sir itis really hard . But I might find a different$ $way .$ $For instance let find the 3 digits numbers in between$ $100 and 200 whose digit sum is 12$ $Sum of digits Numbers$ $1 - - - - 100$ $2 - - - - 101 - - - - 110$ $3 - - - - 102 - - - - 111 - - - - 120$ $4 - - - - 103 - - - - 112 - - - - 121 - - - - 130$ $5 - - - - 104 - - - - 113 - - - - 122 - - - - 131$ $6 - - - - 105 - - - - 114 - - - - 123 - - - - 132$ $7 - - - - 106 - - - - 115 - - - - 124 - - - - 133$ $8 - - - - 107 - - - - 116 - - - - 125 - - - - 134$ $9 - - - - 108 - - - - 117 - - - - 126 - - - - 135$ $10 - - - 109 - - - - 118 - - - - 127 - - - - 136$ $11 - - - - - - - - - 119 - - - - 128 - - - - 137$ $12 - - - - - - - - - - - - - - - 129 - - - - 138$ $and so on . .$ $so the first number is 129 and the last number$ $is 129 + 9 \times n < 200$ $n = 7$ $∴ no . of 3 digits number whose sum is 12 is$ $= 7 + 1 = 8$ $sir is it can be helpful for such problems ?$ $Sir Rasheed and Mr . Wsir please share your$ $comments .$
Commented by mr W last updated on 11/Jul/20

$y e s! t h i s i s t h e b e s t m e t h o d t o s o l v e .$ $p l e a s e r e c h e c k y o u r f o r m u l a, i t h i n k$ $i t c o n t a i n s a n e r r o r . i g o t :$ $C_{5}^{17} - C_{5}^{8} - 5 C_{5}^{7} = 6027$
Commented by mr W last updated on 11/Jul/20

$b i g t h a n k s t o a l l f o r t h e m a n y i d e a s!$
Commented by prakash jain last updated on 11/Jul/20

$See Q 22040$
Commented by prakash jain last updated on 11/Jul/20

$Q 55502$
Commented by prakash jain last updated on 11/Jul/20

$Several other questions answered$ $by mr W only using method .$ $mr W, is the same method not$ $applicable here .$ $I did remember some previous$ $questions so searched .$
Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

$Thank you sirs$
Commented by mr W last updated on 11/Jul/20

$t h e r e i s n o s t a n d a r d w a y f o r s u c h$ $p r o b l e m s . t h e w a y y o u a r e t r y i n g c a n$ $r e a c h t h e g o a l, b u t i t c o u l d b e t o u g h .$
Commented by Rasheed.Sindhi last updated on 11/Jul/20

$S i r a n i d e a c o m e s t o m e! I f s u m$ $o f d i g i t s i s 13 t h a t m e a n s$ $n u m b e r s w h i c h l e a v e s$ $r e m a i n d e r 1 w h e n {t h e y}^{'} r e$ $d i v i d e d b y 3 . T h a t i s 3 k + 1 t y p e$ $n u m b e r s .$ $T h e n u m b e r s l e a v e s r e m a i n d e r$ $4 w h e n {t h e y}^{'} r e d i v i d e d b y 9 .$ $T h a t i s t h e n u m b e r s o f 9 k + 4$ $t y p e s . . . . S o m e e x t r a n u m b e r s$ $. . . .$
Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

$sir but the number$ $300001 = 3 \times 100000 + 1$ $but 3 + 0 + 0 + 0 + 0 + 1 \neq 13$ $and$ $3 \times 94528 + 1 = 283585 \neq 13$
Commented by Rasheed.Sindhi last updated on 11/Jul/20

$H e l l o s i r P R I T H W I S H S E N!$ ${Y o u}^{'} r e r i g h t s i r! S o m e e x t r a$ $n u m b e r s w i l l a l s o i n c l u d e d . S o . . .$
Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

$Please check this$ $1^{st} Digit No . of numbers$ $9 The coefficient of 4 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $8 The coeffi . of 5 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $7 The coeffi . of 6 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $6 The coeffi . of 7 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $5 The coeffi . of 8 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $4 The coeffi . of 9 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $3 The coeffi . of 10 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $2 The coeffi . of 11 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$ $1 The coeffi . of 12 in$ ${(a^{0} + a^{1} + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + a^{7} + a^{8} + a^{9})}^{5}$
Commented by prakash jain last updated on 11/Jul/20

$First digits 1 to 9$ $rest all 0 to 9$ $(x^{1} + . . . . + x^{9}) {(1 + . . . + x^{9})}^{5}$ $Digits with sum 13 is given by$ $coefficient of x^{13} in above expression .$ $\frac{x (1 - x^{9})}{1 - x} \times \frac{{(1 - x^{10})}^{5}}{{(1 - x)}^{5}}$ $= x (1 - x^{9} - x^{10} + higherpower) {(1 - x)}^{- 6}$ $= x (1 - x^{9} - 5 x^{10}) (1 + \underset{i = 1}{\sum^{\infty}}^{6 + i - 1} C_{i} x^{i})$ $= (x - x^{10} - x^{11}) (. . .)$ $Terms in () that will given x^{13}$ $x^{12}, x^{3}, x^{2}$ $^{17} C_{5} -^{8} C_{3} - 5^{7} C_{2}$ $(correction {(1 - x^{10})}^{5} was$ $earlier written (1 - x^{10} + . .)$ $it should be (1 - 5 x^{10} + . .)$
Commented by mr W last updated on 11/Jul/20

$y e s s i r! i s o l v e d s i m i l a r q u e s t i o n s$ $s o m e t i m e s a g o, b u t i {c a n}^{'} t r e m e m b e r$ $t h e o l d p o s t s . h o w d i d y o u f i n d t h e s e$ $o l d p o s t s ?$
Commented by prakash jain last updated on 11/Jul/20

$I searched for generating or just$ $generat etc . tried 2 - 3 search text .$
Commented by mr W last updated on 11/Jul/20

$t h a n k s s i r! i n t h i s w a y o n e c a n f i n d$ $s o m e o l d p o s t s, g r e a t!$
Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

$sir prakash jain$ $i think your 2^{nd} expression will be$ ${(1 + . . . . + x^{9})}^{5} not 6$
Commented by prakash jain last updated on 11/Jul/20

$y e s . i t i s 5 .$
Commented by prakash jain last updated on 11/Jul/20

$o k . I r e a l i z e n o w . i n p r e v i o u s s t e p$ $i h a v e w r i t t e n s i x .$
	Answered by mr W last updated on 11/Jul/20

	${l e t}^{'} s l o o k a t t h e g e n e r a l c a s e :$ $h o w m a n y n d i g i t n u m b e r s e x i s t$ $w h o s e d i g i t s h a v e e x a c t l y t h e s u m m ?$ $s a y s u c h a n u m b e r i s$ $d_{1} d_{2} d_{3} . . . d_{n}$ $w i t h t h e c o n d i t i o n s :$ $1 ⩽ d_{1} ⩽ 9$ $0 ⩽ d_{2}, d_{3}, . . ., d_{n} ⩽ 9$ $s o t h e q u e s t i o n i s h o w m a n y i n t e g r a l$ $s o l u t i o n s t h e f o l l o w i n g e q u a t i o n$ $d_{1} + d_{2} + d_{3} + . . . + d_{n} = m . . . (I)$ $h a s u n d e r t h e g i v e n c o n d i t i o n s .$ $w e c a n u s e t h e g e n e r a t i n g f u n c t i o n s$ $f o r d_{1} : x + x^{2} + x^{3} + . . . + x^{9}$ $f o r d_{2}, d_{3}, . . ., d_{n} : 1 + x + x^{2} + x^{3} + . . . + x^{9}$ $f o r d_{1} + d_{2} + d_{3} + . . . + d_{n} i t i s t h e n$ $(x + x^{2} + x^{3} + . . . + x^{9}) {(1 + x + x^{2} + x^{3} + . . . + x^{9})}^{n - 1}$ $= \frac{x (1 - x^{9}) {(1 - x^{10})}^{n - 1}}{{(1 - x)}^{n}}$ $= x (1 - x^{9}) {(1 - x^{10})}^{n - 1} \underset{k = 0}{\sum^{\infty}} C_{n - 1}^{k + n - 1} x^{k}$ $= x (1 - x^{9}) [\underset{r = 0}{\sum^{n - 1}} {(- 1)}^{r} C_{r}^{n - 1} x^{10 r}] [\underset{k = 0}{\sum^{\infty}} C_{n - 1}^{k + n - 1} x^{k}]$ $t h e c o e f f i c i e n t o f t h e x^{m} t e r m i n t h i s$ $g e n e r a t i n g f u n c t i o n r e p r e s e n t s t h e$ $n u m b e r o f s o l u t i o n s o f e q n . (I) .$ $e x a m p l e : n = 6, m = 13$ $G F = x (1 - x^{9}) (1 - 5 x^{10} + . . .) \underset{k = 0}{\sum^{\infty}} C_{5}^{k + 5} x^{k}$ $= x (1 - x^{9} - 5 x^{10} + . . .) \underset{k = 0}{\sum^{\infty}} C_{5}^{k + 5} x^{k}$ $c o e f f i c i e n t o f x^{13} t e r m i s$ $(f o r k = 12, 3, 2)$ $C_{5}^{17} - C_{5}^{8} - 5 C_{5}^{7} = 6027$
	Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

	$Thanks sir$
	Answered by prakash jain last updated on 11/Jul/20

	$x x x x x x x x x x x x x x x x x x (18 x)$ $Replace any 5 of last 17 x by a bar$ $count the number of x s between$ $to bars to get a digit .$ $Now this will include cases where$ $where all 5 bars are included in$ $8 or less continous x s . Meaning digits$ $has to be ⩽ 9 .$ $You can't use 'macro parameter character #' in math mode$ $one digits is 10 = 6 \times^{5} C_{3} +^{7} C_{4} +^{6} C_{4} = 110$ $one digits is 11 = 5 \times^{4} C_{3} +^{6} C_{4} +^{5} C_{4} = 55$ $one digits is 12 = 4 \times^{3} C_{3} +^{5} C_{4} +^{4} C_{4} = 10$ $one digits is 13 = 1$ $valid outcomes$ $=^{17} C_{5} - 161 = 6188 - 161 = 6027$ $count the number of x between$ $2 bars to get the digit .$ $x ∣ 10 x ∣ 5 x to x 6 x ∣ 10 x$ $x x ∣ x x x ∣ x x x ∣∣ x x x ∣ x x$ $233032$ $x x ∣∣∣∣∣∣ x x x x x x x x x x x$
	Commented by Rasheed.Sindhi last updated on 11/Jul/20

	$S i r r e a l l y u s e f u l m e t h o d! I$ $r e m e m b e r n o w t h a t l o n g a g o I$ $h a d l e a r n t i t f r o m y o u b u t I$ $f o r g o t!$
	Commented by prakash jain last updated on 11/Jul/20

	$This method is commonly known$ $as stars and bars .$ $Very useful for dividing n items$ $in m boxes .$ $The above is a special cases where$ $first box is not empty and other$ $boxes can be empty but cannot$ $contain more than 9 items .$
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