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Question Number 117281 by mr W last updated on 10/Oct/20

I n h o w m a n y d i f f e r e n t w a y s c a n w e

s e l e c t 5 n u m b e r s f r o m 9 n u m b e r s

{1, 2, 3, \dots, 9} ?

A n u m b e r m a y b e s e l e c t e d m o r e t h a n

o n e t i m e .

Answered by PRITHWISH SEN 2 last updated on 10/Oct/20

think as all the latice point in a 5^{th} dimensional

coordinate systems with only 9 points in each

axes

that is in 9^{5} ways

Commented by mr W last updated on 10/Oct/20

t h i s i s a l s o m y a n s w e r .

w e c a n a r r a n g e e a c h g r o u p o f 5 d i g i t s

i n i n c r e a s i n g o r d e r t o f o r m a 5 d i g i t

n u m b e r x_{1} x_{2} x_{3} x_{4} x_{5} w i t h

x_{1} ⩽ x_{2} ⩽ x_{3} ⩽ x_{4} ⩽ x_{5}

i n t h i s w a y i g o t t h e a n s w e r 9^{5} = 59049

t o Q 117061 .

w h a t i s w r o n g i n m y t h i n k i n g ?

Commented by PRITHWISH SEN 2 last updated on 10/Oct/20

no sir yours is fine . I just want to visualize .

Is it ok ?

Commented by mr W last updated on 10/Oct/20

t h a n k s s i r!

Commented by MJS_new last updated on 11/Oct/20

if you were right in the case with 5 digits

then what would be the answer for 2, 3, 4, n

digits ? 9^{n} is wrong for 2 digits :

x_{1} x_{2} with x_{1} ⩽ x_{2}

11 12 13 14 15 16 17 18 19

22 23 24 25 26 27 28 29

33 34 35 36 37 38 39

44 45 46 47 48 49

55 56 57 58 59

66 67 68 69

77 78 79

88 89

99

9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 \neq 9^{2}

Commented by mr W last updated on 11/Oct/20

i a l s o r e a l i s e d t h a t i t i s n o t c o r r e c t .

b u t i {d o n}^{'} t s e e w h e r e i s w r o n g i n m y

t h i n k i n g :

f r o m 9 d i g i t s w e h a v e 9^{n} w a y s t o

s e l e c t n d i g i t s, i n e a c h o f t h e s e w a y s

w e c a n f o r m a v a l i d n u m b e r f u l f i l l i n g

x_{1} ⩽ x_{2} ⩽ \dots ⩽ x_{n} . t h a t m e a n s t h e r e

s h o u l d b e 9^{n} s u c h n u m b e r s .

Commented by MJS_new last updated on 11/Oct/20

staying with 2 digits

the pairs (x_{1}, x_{2}) and (x_{2}, x_{1}) with x_{1} < x_{2}

form the same number x_{1} x_{2}

with more digits {it}^{'} s getting more complicated

3 digits i . e .

(1, 2, 3) has 5 “ {siblings}^{″} but (1, 1, 2) only 3

\dots

Commented by mr W last updated on 11/Oct/20

(x_{1}, x_{2}) and (x_{2}, x_{1}) s h o u l d b e s e e n a s

t h e s a m e s e l e c t i o n .

Commented by mr W last updated on 11/Oct/20

n o w i s e e t h e a n s w e r 9^{n} i s w r o n g .

w e h a v e 9 d i g i t s . e a c h o f t h e m m a y

b e s e l e c t e d 0, 1, 2, \dots t i m e s . t o t a l l y

n d i g i t s s h o u l d b e s e l e c t e d . t h e n u m b e r

o f w a y s t o d o t h i s i s t h e c o e f f i c i e n t

o f x^{n} t e r m i n \frac{1}{{(1 - x)}^{9}} . t h a t i s C_{8}^{n + 8} .

s o t h e c o r r e c t a n s w e r i s C_{8}^{n + 8} .

C_{8}^{n + 8} = \frac{(n + 8)!}{8! n!}

Commented by mr W last updated on 11/Oct/20

Commented by PRITHWISH SEN 2 last updated on 11/Oct/20

no such condition as x_{1} ⩽ x_{2} \dots is given for this sum .

so i think in this case 9^{5} is correct answer .

Commented by mr W last updated on 11/Oct/20

i t h i n k 9^{5} i s n o t c o r r e c t, f o r e x a m p l e,

12345

13452

54321

a r e t r e a t e d a s d i f f e r e n t w a y s,

b u t i n f a c t t h e y a r e t h e s a m e w a y .

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