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Question Number 117281 by mr W last updated on 10/Oct/20
In how many different ways can we  select 5 numbers from 9 numbers  {1,2,3,...,9}?   A number may be selected more than  one time.
$${In}\:{how}\:{many}\:{different}\:{ways}\:{can}\:{we} \\ $$$${select}\:\mathrm{5}\:{numbers}\:{from}\:\mathrm{9}\:{numbers} \\ $$$$\left\{\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{9}\right\}?\: \\ $$$${A}\:{number}\:{may}\:{be}\:{selected}\:{more}\:{than} \\ $$$${one}\:{time}. \\ $$
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
$$\mathrm{think}\:\mathrm{as}\:\mathrm{all}\:\mathrm{the}\:\mathrm{latice}\:\mathrm{point}\:\mathrm{in}\:\mathrm{a}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{dimensional} \\ $$$$\mathrm{coordinate}\:\mathrm{systems}\:\mathrm{with}\:\mathrm{only}\:\mathrm{9}\:\mathrm{points}\:\mathrm{in}\:\mathrm{each} \\ $$$$\mathrm{axes} \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{in}\:\mathrm{9}^{\mathrm{5}} \mathrm{ways} \\ $$
Commented by mr W last updated on 10/Oct/20
this is also my answer.  we can arrange each group of 5 digits  in increasing order to form a 5 digit  number x_1 x_2 x_3 x_4 x_5  with  x_1 ≤x_2 ≤x_3 ≤x_4 ≤x_5   in this way i got the answer 9^5 =59049  to Q117061.  what is wrong in my thinking?
$${this}\:{is}\:{also}\:{my}\:{answer}. \\ $$$${we}\:{can}\:{arrange}\:{each}\:{group}\:{of}\:\mathrm{5}\:{digits} \\ $$$${in}\:{increasing}\:{order}\:{to}\:{form}\:{a}\:\mathrm{5}\:{digit} \\ $$$${number}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} {x}_{\mathrm{4}} {x}_{\mathrm{5}} \:{with} \\ $$$${x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \leqslant{x}_{\mathrm{3}} \leqslant{x}_{\mathrm{4}} \leqslant{x}_{\mathrm{5}} \\ $$$${in}\:{this}\:{way}\:{i}\:{got}\:{the}\:{answer}\:\mathrm{9}^{\mathrm{5}} =\mathrm{59049} \\ $$$${to}\:{Q}\mathrm{117061}. \\ $$$${what}\:{is}\:{wrong}\:{in}\:{my}\:{thinking}? \\ $$
Commented by PRITHWISH SEN 2 last updated on 10/Oct/20
no sir yours is fine. I just want to visualize.  Is it ok ?
$$\mathrm{no}\:\mathrm{sir}\:\mathrm{yours}\:\mathrm{is}\:\mathrm{fine}.\:\mathrm{I}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{visualize}. \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{ok}\:? \\ $$
Commented by mr W last updated on 10/Oct/20
thanks sir!
$${thanks}\:{sir}! \\ $$
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 ≠ 9^2
$$\mathrm{if}\:\mathrm{you}\:\mathrm{were}\:\mathrm{right}\:\mathrm{in}\:\mathrm{the}\:\mathrm{case}\:\mathrm{with}\:\mathrm{5}\:\mathrm{digits} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{would}\:\mathrm{be}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{for}\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:{n} \\ $$$$\mathrm{digits}?\:\mathrm{9}^{{n}} \:\mathrm{is}\:\mathrm{wrong}\:\mathrm{for}\:\mathrm{2}\:\mathrm{digits}: \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} \:\mathrm{with}\:{x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \\ $$$$\mathrm{11}\:\mathrm{12}\:\mathrm{13}\:\mathrm{14}\:\mathrm{15}\:\mathrm{16}\:\mathrm{17}\:\mathrm{18}\:\mathrm{19} \\ $$$$\mathrm{22}\:\mathrm{23}\:\mathrm{24}\:\mathrm{25}\:\mathrm{26}\:\mathrm{27}\:\mathrm{28}\:\mathrm{29} \\ $$$$\mathrm{33}\:\mathrm{34}\:\mathrm{35}\:\mathrm{36}\:\mathrm{37}\:\mathrm{38}\:\mathrm{39} \\ $$$$\mathrm{44}\:\mathrm{45}\:\mathrm{46}\:\mathrm{47}\:\mathrm{48}\:\mathrm{49} \\ $$$$\mathrm{55}\:\mathrm{56}\:\mathrm{57}\:\mathrm{58}\:\mathrm{59} \\ $$$$\mathrm{66}\:\mathrm{67}\:\mathrm{68}\:\mathrm{69} \\ $$$$\mathrm{77}\:\mathrm{78}\:\mathrm{79} \\ $$$$\mathrm{88}\:\mathrm{89} \\ $$$$\mathrm{99} \\ $$$$\mathrm{9}+\mathrm{8}+\mathrm{7}+\mathrm{6}+\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}=\mathrm{45}\:\neq\:\mathrm{9}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 11/Oct/20
i also realised that it is not correct.  but i don′t see where is wrong in my  thinking:  from 9 digits we have 9^n  ways to  select n digits, in each of these ways  we can form a valid number fulfilling  x_1 ≤x_2 ≤...≤x_n . that means there  should be 9^n  such numbers.
$${i}\:{also}\:{realised}\:{that}\:{it}\:{is}\:{not}\:{correct}. \\ $$$${but}\:{i}\:{don}'{t}\:{see}\:{where}\:{is}\:{wrong}\:{in}\:{my} \\ $$$${thinking}: \\ $$$${from}\:\mathrm{9}\:{digits}\:{we}\:{have}\:\mathrm{9}^{{n}} \:{ways}\:{to} \\ $$$${select}\:{n}\:{digits},\:{in}\:{each}\:{of}\:{these}\:{ways} \\ $$$${we}\:{can}\:{form}\:{a}\:{valid}\:{number}\:{fulfilling} \\ $$$${x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \leqslant…\leqslant{x}_{{n}} .\:{that}\:{means}\:{there} \\ $$$${should}\:{be}\:\mathrm{9}^{{n}} \:{such}\:{numbers}. \\ $$
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  ...
$$\mathrm{staying}\:\mathrm{with}\:\mathrm{2}\:\mathrm{digits} \\ $$$$\mathrm{the}\:\mathrm{pairs}\:\left({x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} \right)\:\mathrm{and}\:\left({x}_{\mathrm{2}} ,\:{x}_{\mathrm{1}} \right)\:\mathrm{with}\:{x}_{\mathrm{1}} <{x}_{\mathrm{2}} \\ $$$$\mathrm{form}\:\mathrm{the}\:\mathrm{same}\:\mathrm{number}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\mathrm{with}\:\mathrm{more}\:\mathrm{digits}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting}\:\mathrm{more}\:\mathrm{complicated} \\ $$$$\mathrm{3}\:\mathrm{digits}\:\mathrm{i}.\mathrm{e}. \\ $$$$\left(\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right)\:\mathrm{has}\:\mathrm{5}\:“\mathrm{siblings}''\mathrm{but}\:\left(\mathrm{1},\:\mathrm{1},\:\mathrm{2}\right)\:\mathrm{only}\:\mathrm{3} \\ $$$$… \\ $$
Commented by mr W last updated on 11/Oct/20
(x_1 , x_2 ) and (x_2 , x_1 ) should be seen as  the same selection.
$$\left({x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} \right)\:\mathrm{and}\:\left({x}_{\mathrm{2}} ,\:{x}_{\mathrm{1}} \right)\:{should}\:{be}\:{seen}\:{as} \\ $$$${the}\:{same}\:{selection}. \\ $$
Commented by mr W last updated on 11/Oct/20
now i see the answer 9^n  is wrong.  we have 9 digits. each of them may  be selected 0,1,2,... times. totally  n digits should be selected. the number  of ways to do this is the coefficient  of x^n  term in (1/((1−x)^9 )). that is C_8 ^(n+8) .  so the correct answer is C_8 ^(n+8) .  C_8 ^(n+8) =(((n+8)!)/(8!n!))
$${now}\:{i}\:{see}\:{the}\:{answer}\:\mathrm{9}^{{n}} \:{is}\:{wrong}. \\ $$$${we}\:{have}\:\mathrm{9}\:{digits}.\:{each}\:{of}\:{them}\:{may} \\ $$$${be}\:{selected}\:\mathrm{0},\mathrm{1},\mathrm{2},…\:{times}.\:{totally} \\ $$$${n}\:{digits}\:{should}\:{be}\:{selected}.\:{the}\:{number} \\ $$$${of}\:{ways}\:{to}\:{do}\:{this}\:{is}\:{the}\:{coefficient} \\ $$$${of}\:{x}^{{n}} \:{term}\:{in}\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{9}} }.\:{that}\:{is}\:{C}_{\mathrm{8}} ^{{n}+\mathrm{8}} . \\ $$$${so}\:{the}\:{correct}\:{answer}\:{is}\:{C}_{\mathrm{8}} ^{{n}+\mathrm{8}} . \\ $$$${C}_{\mathrm{8}} ^{{n}+\mathrm{8}} =\frac{\left({n}+\mathrm{8}\right)!}{\mathrm{8}!{n}!} \\ $$
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 ... is given for this sum.  so i think in this case 9^5  is correct answer.
$$\mathrm{no}\:\mathrm{such}\:\mathrm{condition}\:\mathrm{as}\:\mathrm{x}_{\mathrm{1}} \leqslant\mathrm{x}_{\mathrm{2}} …\:\mathrm{is}\:\mathrm{given}\:\mathrm{for}\:\mathrm{this}\:\mathrm{sum}. \\ $$$$\mathrm{so}\:\mathrm{i}\:\mathrm{think}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{9}^{\mathrm{5}} \:\mathrm{is}\:\mathrm{correct}\:\mathrm{answer}. \\ $$
Commented by mr W last updated on 11/Oct/20
i think 9^5  is not correct, for example,  12345  13452  54321  are treated as different ways,  but in fact they are the same way.
$${i}\:{think}\:\mathrm{9}^{\mathrm{5}} \:{is}\:{not}\:{correct},\:{for}\:{example}, \\ $$$$\mathrm{12345} \\ $$$$\mathrm{13452} \\ $$$$\mathrm{54321} \\ $$$${are}\:{treated}\:{as}\:{different}\:{ways}, \\ $$$${but}\:{in}\:{fact}\:{they}\:{are}\:{the}\:{same}\:{way}. \\ $$

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