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Let-S-1-2-3-4-5-we-want-to-make-a-group-H-of-numbers-have-the-following-properties-1-Each-number-has-different-digits-and-taken-from-S-2-Each-number-is-greater-than-20-000-3-None-




Question Number 178021 by Acem last updated on 12/Oct/22
  Let S={1, 2, 3, 4, 5} , we want to make a    group H of numbers have the following    properties:   1. Each number has different digits and   taken from S   2. Each number is greater than 20 000   3. None of them is multiple of five    How many items of H?
$$ \\ $$$${Let}\:{S}=\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right\}\:,\:{we}\:{want}\:{to}\:{make}\:{a}\: \\ $$$$\:{group}\:{H}\:{of}\:{numbers}\:{have}\:{the}\:{following}\: \\ $$$$\:{properties}: \\ $$$$\:\mathrm{1}.\:{Each}\:{number}\:{has}\:{different}\:{digits}\:{and} \\ $$$$\:{taken}\:{from}\:{S} \\ $$$$\:\mathrm{2}.\:{Each}\:{number}\:{is}\:{greater}\:{than}\:\mathrm{20}\:\mathrm{000} \\ $$$$\:\mathrm{3}.\:{None}\:{of}\:{them}\:{is}\:{multiple}\:{of}\:{five} \\ $$$$ \\ $$$${How}\:{many}\:{items}\:{of}\:{H}? \\ $$
Answered by mr W last updated on 12/Oct/22
4×4×3×2×1−3×3×2×1=78
$$\mathrm{4}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}−\mathrm{3}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{78} \\ $$
Commented by Tawa11 last updated on 12/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by aurpeyz last updated on 12/Oct/22
thank you Sir. I got how you handled the  first two but i didnt gwt how you did that  of the multiples of 5. pls explain
$${thank}\:{you}\:{Sir}.\:{I}\:{got}\:{how}\:{you}\:{handled}\:{the} \\ $$$${first}\:{two}\:{but}\:{i}\:{didnt}\:{gwt}\:{how}\:{you}\:{did}\:{that} \\ $$$${of}\:{the}\:{multiples}\:{of}\:\mathrm{5}.\:{pls}\:{explain}\: \\ $$
Commented by mr W last updated on 12/Oct/22
let′s begin with the last digit, which  should not be 5. so we have 4 ways to  select the last digit. for the second  last digit we have also 4 possiblities  and for the next digit 3 possibilities  etc. so tatally we have 4×4×3×2×1  ways.  among them some have 1 as  the first digit. these numbers are  not valid, because they are smaller  than 20000. now we shoud find how  mang numbers begin with 1.  1abcd  abcd are from 2,3,4,5. similary as  above we can get that there are  3×3×2×1 such numbers. so the  answer is  4×4×3×2×1−3×3×2×1=78.
$${let}'{s}\:{begin}\:{with}\:{the}\:{last}\:{digit},\:{which} \\ $$$${should}\:{not}\:{be}\:\mathrm{5}.\:{so}\:{we}\:{have}\:\mathrm{4}\:{ways}\:{to} \\ $$$${select}\:{the}\:{last}\:{digit}.\:{for}\:{the}\:{second} \\ $$$${last}\:{digit}\:{we}\:{have}\:{also}\:\mathrm{4}\:{possiblities} \\ $$$${and}\:{for}\:{the}\:{next}\:{digit}\:\mathrm{3}\:{possibilities} \\ $$$${etc}.\:{so}\:{tatally}\:{we}\:{have}\:\mathrm{4}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1} \\ $$$${ways}.\:\:{among}\:{them}\:{some}\:{have}\:\mathrm{1}\:{as} \\ $$$${the}\:{first}\:{digit}.\:{these}\:{numbers}\:{are} \\ $$$${not}\:{valid},\:{because}\:{they}\:{are}\:{smaller} \\ $$$${than}\:\mathrm{20000}.\:{now}\:{we}\:{shoud}\:{find}\:{how} \\ $$$${mang}\:{numbers}\:{begin}\:{with}\:\mathrm{1}. \\ $$$$\mathrm{1}{abcd} \\ $$$${abcd}\:{are}\:{from}\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}.\:{similary}\:{as} \\ $$$${above}\:{we}\:{can}\:{get}\:{that}\:{there}\:{are} \\ $$$$\mathrm{3}×\mathrm{3}×\mathrm{2}×\mathrm{1}\:{such}\:{numbers}.\:{so}\:{the} \\ $$$${answer}\:{is} \\ $$$$\mathrm{4}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}−\mathrm{3}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{78}. \\ $$
Commented by Acem last updated on 12/Oct/22
I appreciate the professor′s method, it′s very   clear and easy. I was going to mention mine   but it′s a bit complicated.  Big thanks
$${I}\:{appreciate}\:{the}\:{professor}'{s}\:{method},\:{it}'{s}\:{very} \\ $$$$\:{clear}\:{and}\:{easy}.\:{I}\:{was}\:{going}\:{to}\:{mention}\:{mine} \\ $$$$\:{but}\:{it}'{s}\:{a}\:{bit}\:{complicated}. \\ $$$$\boldsymbol{{Big}}\:\boldsymbol{{thanks}} \\ $$
Commented by aurpeyz last updated on 12/Oct/22
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$
Commented by Acem last updated on 12/Oct/22
 A simple explanation of the almighty   professor′s method                                                         a= 20 000    determinant ((,,,,(Not 5),(NW_(Num_s >a or Num_s <a) = 96)),(1,2,3,4,(    4),(← Start from here)))      determinant ((( 1),,,,(Not 5),(NW_(Num_s < a)  = 18)),(−,1,2,3,(    3),(← Start from here)))     NumItems_H  = the difference= 78 numbers
$$\:{A}\:{simple}\:{explanation}\:{of}\:{the}\:{almighty} \\ $$$$\:{professor}'{s}\:{method} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\:\mathrm{20}\:\mathrm{000} \\ $$$$\:\begin{array}{|c|c|}{}&\hline{}&\hline{}&\hline{}&\hline{{Not}\:\mathrm{5}}&\hline{{NW}_{{Num}_{{s}} >{a}\:{or}\:{Num}_{{s}} <{a}} =\:\mathrm{96}}\\{\mathrm{1}}&\hline{\mathrm{2}}&\hline{\mathrm{3}}&\hline{\mathrm{4}}&\hline{\:\:\:\:\mathrm{4}}&\hline{\leftarrow\:{Start}\:{from}\:{here}}\\\hline\end{array} \\ $$$$ \\ $$$$\:\begin{array}{|c|c|}{\:\mathrm{1}}&\hline{}&\hline{}&\hline{}&\hline{{Not}\:\mathrm{5}}&\hline{{NW}_{{Num}_{{s}} <\:{a}} \:=\:\mathrm{18}}\\{−}&\hline{\mathrm{1}}&\hline{\mathrm{2}}&\hline{\mathrm{3}}&\hline{\:\:\:\:\mathrm{3}}&\hline{\leftarrow\:{Start}\:{from}\:{here}}\\\hline\end{array} \\ $$$$ \\ $$$$\:{NumItems}_{{H}} \:=\:{the}\:{difference}=\:\mathrm{78}\:{numbers} \\ $$$$ \\ $$
Commented by aurpeyz last updated on 12/Oct/22
nice one
$${nice}\:{one} \\ $$

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