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-

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

$${In}\:{how}\:{many}\:{different}\:{ways}\:{can}\:{we} \\ $$$${select}\:\mathrm{5}\:{numbers}\:{from}\:\mathrm{9}\:{numbers} \\ $$$$\left\{\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{9}\right\}?\: \\ $$$${A}\:{number}\:{may}\:{be}\:{selected}\:{more}\:{than} \\ $$$${one}\:{time}. \\ $$

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

$${thanks}\:{sir}! \\ $$

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

$$\mathrm{if}\:\mathrm{you}\:\mathrm{were}\:\mathrm{right}\:\mathrm{in}\:\mathrm{the}\:\mathrm{case}\:\mathrm{with}\:\mathrm{5}\:\mathrm{digits} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{would}\:\mathrm{be}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{for}\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:{n} \\ $$$$\mathrm{digits}?\:\mathrm{9}^{{n}} \:\mathrm{is}\:\mathrm{wrong}\:\mathrm{for}\:\mathrm{2}\:\mathrm{digits}: \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} \:\mathrm{with}\:{x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \\ $$$$\mathrm{11}\:\mathrm{12}\:\mathrm{13}\:\mathrm{14}\:\mathrm{15}\:\mathrm{16}\:\mathrm{17}\:\mathrm{18}\:\mathrm{19} \\ $$$$\mathrm{22}\:\mathrm{23}\:\mathrm{24}\:\mathrm{25}\:\mathrm{26}\:\mathrm{27}\:\mathrm{28}\:\mathrm{29} \\ $$$$\mathrm{33}\:\mathrm{34}\:\mathrm{35}\:\mathrm{36}\:\mathrm{37}\:\mathrm{38}\:\mathrm{39} \\ $$$$\mathrm{44}\:\mathrm{45}\:\mathrm{46}\:\mathrm{47}\:\mathrm{48}\:\mathrm{49} \\ $$$$\mathrm{55}\:\mathrm{56}\:\mathrm{57}\:\mathrm{58}\:\mathrm{59} \\ $$$$\mathrm{66}\:\mathrm{67}\:\mathrm{68}\:\mathrm{69} \\ $$$$\mathrm{77}\:\mathrm{78}\:\mathrm{79} \\ $$$$\mathrm{88}\:\mathrm{89} \\ $$$$\mathrm{99} \\ $$$$\mathrm{9}+\mathrm{8}+\mathrm{7}+\mathrm{6}+\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}=\mathrm{45}\:\neq\:\mathrm{9}^{\mathrm{2}} \\ $$