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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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