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Question Number 178134 by Acem last updated on 13/Oct/22

Am not a friend with this isssue:  6 red, 1 black and 3 white balls  We draw 3 balls in a raw, returning the   drawn ball each time.  How many different results which include   at least one black ball.     Way_1 : 1000−9^3 = 271 results That′s very ok     Way_2 : 1×9×9_(very ok) ×3_(Omg, Why?)                  + 1×1×9×3_(the same confused above)                  + 1×1×1_(very ok and i like it.... but why not ×3)                   = 271 result     Explanation_(about the story of ×3) ?

$${Am}\:{not}\:{a}\:{friend}\:{with}\:{this}\:{isssue}: \\ $$$$\mathrm{6}\:{red},\:\mathrm{1}\:{black}\:{and}\:\mathrm{3}\:{white}\:{balls} \\ $$$${We}\:{draw}\:\mathrm{3}\:{balls}\:{in}\:{a}\:{raw},\:{returning}\:{the} \\ $$$$\:{drawn}\:{ball}\:{each}\:{time}. \\ $$$${How}\:{many}\:{different}\:{results}\:{which}\:{include} \\ $$$$\:\boldsymbol{{at}}\:\boldsymbol{{least}}\:{one}\:{black}\:{ball}. \\ $$$$ \\ $$$$\:{Way}_{\mathrm{1}} :\:\mathrm{1000}−\mathrm{9}^{\mathrm{3}} =\:\mathrm{271}\:{results}\:{That}'{s}\:{very}\:{ok} \\ $$$$ \\ $$$$\:{Way}_{\mathrm{2}} :\:\mathrm{1}×\mathrm{9}×\mathrm{9}_{{very}\:{ok}} ×\mathrm{3}_{\boldsymbol{{Omg}},\:\boldsymbol{{Why}}?} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{1}×\mathrm{1}×\mathrm{9}×\mathrm{3}_{{the}\:{same}\:{confused}\:{above}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{1}×\mathrm{1}×\mathrm{1}_{{very}\:{ok}\:{and}\:{i}\:{like}\:{it}....\:{but}\:{why}\:{not}\:×\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{271}\:{result} \\ $$$$ \\ $$$$\:{Explanation}_{{about}\:{the}\:{story}\:{of}\:×\mathrm{3}} ? \\ $$$$ \\ $$

Commented by Acem last updated on 13/Oct/22

 He asked me about final getting 1 black ball.   I done! the question doesn′t care   of arranging things.   Anyway give me some time till Freon13   settle in my brain tank.

$$\:{He}\:{asked}\:{me}\:{about}\:{final}\:{getting}\:\mathrm{1}\:{black}\:{ball}. \\ $$$$\:{I}\:{done}!\:{the}\:{question}\:{doesn}'{t}\:{care} \\ $$$$\:{of}\:{arranging}\:{things}. \\ $$$$\:{Anyway}\:{give}\:{me}\:{some}\:{time}\:{till}\:{Freon}\mathrm{13} \\ $$$$\:{settle}\:{in}\:{my}\:{brain}\:{tank}.\: \\ $$$$ \\ $$

Commented by mr W last updated on 13/Oct/22

Explanation to Way 2  case 1: one black ball, two balls from  other colors  1×9×9  since the black ball can be the first one,  the second one or the third one,  therefore totally 1×9×9×3 possibilities  case 2: two black balls and one ball  from other colors  1×1×9  since the black balls can be the first two,  the last two or the first and the last  one,  therefore totally 1×1×9×3   possibilities  case 3: three black balls  1×1×1    all together:  1×9×9×3+1×1×9×3+1×1×1=271

$${Explanation}\:{to}\:{Way}\:\mathrm{2} \\ $$$${case}\:\mathrm{1}:\:{one}\:{black}\:{ball},\:{two}\:{balls}\:{from} \\ $$$${other}\:{colors} \\ $$$$\mathrm{1}×\mathrm{9}×\mathrm{9} \\ $$$${since}\:{the}\:{black}\:{ball}\:{can}\:{be}\:{the}\:{first}\:{one}, \\ $$$${the}\:{second}\:{one}\:{or}\:{the}\:{third}\:{one}, \\ $$$${therefore}\:{totally}\:\mathrm{1}×\mathrm{9}×\mathrm{9}×\mathrm{3}\:{possibilities} \\ $$$${case}\:\mathrm{2}:\:{two}\:{black}\:{balls}\:{and}\:{one}\:{ball} \\ $$$${from}\:{other}\:{colors} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{9} \\ $$$${since}\:{the}\:{black}\:{balls}\:{can}\:{be}\:{the}\:{first}\:{two}, \\ $$$${the}\:{last}\:{two}\:{or}\:{the}\:{first}\:{and}\:{the}\:{last} \\ $$$${one},\:\:{therefore}\:{totally}\:\mathrm{1}×\mathrm{1}×\mathrm{9}×\mathrm{3}\: \\ $$$${possibilities} \\ $$$${case}\:\mathrm{3}:\:{three}\:{black}\:{balls} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{1} \\ $$$$ \\ $$$${all}\:{together}: \\ $$$$\mathrm{1}×\mathrm{9}×\mathrm{9}×\mathrm{3}+\mathrm{1}×\mathrm{1}×\mathrm{9}×\mathrm{3}+\mathrm{1}×\mathrm{1}×\mathrm{1}=\mathrm{271} \\ $$

Commented by mr W last updated on 13/Oct/22

if we had 2 black balls instead of one  black ball in the bag, the result is  2×9×9×3+2×2×9×3+2×2×2=602  we can get this result also from  11^3 −9^3 =602

$${if}\:{we}\:{had}\:\mathrm{2}\:{black}\:{balls}\:{instead}\:{of}\:{one} \\ $$$${black}\:{ball}\:{in}\:{the}\:{bag},\:{the}\:{result}\:{is} \\ $$$$\mathrm{2}×\mathrm{9}×\mathrm{9}×\mathrm{3}+\mathrm{2}×\mathrm{2}×\mathrm{9}×\mathrm{3}+\mathrm{2}×\mathrm{2}×\mathrm{2}=\mathrm{602} \\ $$$${we}\:{can}\:{get}\:{this}\:{result}\:{also}\:{from} \\ $$$$\mathrm{11}^{\mathrm{3}} −\mathrm{9}^{\mathrm{3}} =\mathrm{602} \\ $$

Commented by Tawa11 last updated on 13/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Acem last updated on 13/Oct/22

  First of all, thank you for your interest and   detailed explanation.    According to mathematical logic, we care about   the order of the elements of the chosen set.  For example choosing letters from a group to  form a word (diff. words by changing the letters),  choosing numbers to form numbers(diff. num.),  choosing people to assign them  specific tasks (diff. person′s job)...etc    In all the above we concered in all   their  possibility permutations   So how can you make me convince or accept   mentally of the importance of the order of the   selected only different color balls:   • • • , • • • Here i see no difference, I can′t   see that there′s different jobs, different phrases   neither different numbers   or even different forms.    The result is i got one, two.. etc black balls     According to way_1 : 10^3 −9^3 = 271   i know am wrong but not in all cases we can use   the way_1  and that time will use the general   principle counting then i′m sure i′ll make   a mistake.     Please help me to see that there are differences   in the permutations at this issue.   And make me friend with it.

$$ \\ $$$${First}\:{of}\:{all},\:{thank}\:{you}\:{for}\:{your}\:{interest}\:{and} \\ $$$$\:{detailed}\:{explanation}. \\ $$$$ \\ $$$${According}\:{to}\:{mathematical}\:{logic},\:{we}\:{care}\:{about} \\ $$$$\:{the}\:{order}\:{of}\:{the}\:{elements}\:{of}\:{the}\:{chosen}\:{set}. \\ $$$${For}\:{example}\:{choosing}\:{letters}\:{from}\:{a}\:{group}\:{to} \\ $$$${form}\:{a}\:{word}\:\left({diff}.\:{words}\:{by}\:{changing}\:{the}\:{letters}\right), \\ $$$${choosing}\:{numbers}\:{to}\:{form}\:{numbers}\left({diff}.\:{num}.\right), \\ $$$${choosing}\:{people}\:{to}\:{assign}\:{them} \\ $$$${specific}\:{tasks}\:\left({diff}.\:{person}'{s}\:{job}\right)...{etc} \\ $$$$ \\ $$$${In}\:{all}\:{the}\:{above}\:{we}\:{concered}\:{in}\:{all} \\ $$$$\:{their}\:\:{possibility}\:{permutations} \\ $$$$\:\boldsymbol{{So}}\:\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{you}}\:\boldsymbol{{make}}\:\boldsymbol{{me}}\:{convince}\:{or}\:{accept} \\ $$$$\:{mentally}\:{of}\:{the}\:{importance}\:{of}\:{the}\:{order}\:{of}\:{the} \\ $$$$\:{selected}\:{only}\:{different}\:{color}\:{balls}: \\ $$$$\:\bullet\:\bullet\:\bullet\:,\:\bullet\:\bullet\:\bullet\:{Here}\:{i}\:{see}\:{no}\:{difference},\:{I}\:{can}'{t} \\ $$$$\:{see}\:{that}\:{there}'{s}\:{different}\:{jobs},\:{different}\:{phrases} \\ $$$$\:{neither}\:{different}\:{numbers} \\ $$$$\:{or}\:{even}\:{different}\:{forms}. \\ $$$$ \\ $$$${The}\:{result}\:{is}\:{i}\:{got}\:{one},\:{two}..\:{etc}\:{black}\:{balls} \\ $$$$ \\ $$$$\:{According}\:{to}\:{way}_{\mathrm{1}} :\:\mathrm{10}^{\mathrm{3}} −\mathrm{9}^{\mathrm{3}} =\:\mathrm{271} \\ $$$$\:{i}\:{know}\:{am}\:{wrong}\:{but}\:{not}\:{in}\:{all}\:{cases}\:{we}\:{can}\:{use} \\ $$$$\:{the}\:{way}_{\mathrm{1}} \:{and}\:{that}\:{time}\:{will}\:{use}\:{the}\:{general} \\ $$$$\:{principle}\:{counting}\:{then}\:{i}'{m}\:{sure}\:{i}'{ll}\:{make} \\ $$$$\:{a}\:{mistake}. \\ $$$$ \\ $$$$\:{Please}\:{help}\:{me}\:{to}\:{see}\:{that}\:{there}\:{are}\:{differences} \\ $$$$\:{in}\:{the}\:{permutations}\:{at}\:{this}\:{issue}. \\ $$$$\:{And}\:{make}\:{me}\:{friend}\:{with}\:{it}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 13/Oct/22

i can understand you doubt.  even when we are not talking about  probability in current question, but  actually it is about probability. then  we must treat the balls with the same  color as different objects. example:  in a bag there are 5 red balls and one  black ball. now you randomly take  two balls. how many possibilities  for taking one red and one black ball?  we know there are 5 possibilities,  even they all look like the same:  black+red. you don′t care which red   ball is taken, because they are the  same to you. but the balls themself  care. there is difference if the ball  A is taken or the ball B is taken.

$${i}\:{can}\:{understand}\:{you}\:{doubt}. \\ $$$${even}\:{when}\:{we}\:{are}\:{not}\:{talking}\:{about} \\ $$$${probability}\:{in}\:{current}\:{question},\:{but} \\ $$$${actually}\:{it}\:{is}\:{about}\:{probability}.\:{then} \\ $$$${we}\:{must}\:{treat}\:{the}\:{balls}\:{with}\:{the}\:{same} \\ $$$${color}\:{as}\:{different}\:{objects}.\:{example}: \\ $$$${in}\:{a}\:{bag}\:{there}\:{are}\:\mathrm{5}\:{red}\:{balls}\:{and}\:{one} \\ $$$${black}\:{ball}.\:{now}\:{you}\:{randomly}\:{take} \\ $$$${two}\:{balls}.\:{how}\:{many}\:{possibilities} \\ $$$${for}\:{taking}\:{one}\:{red}\:{and}\:{one}\:{black}\:{ball}? \\ $$$${we}\:{know}\:{there}\:{are}\:\mathrm{5}\:{possibilities}, \\ $$$${even}\:{they}\:{all}\:{look}\:{like}\:{the}\:{same}: \\ $$$${black}+{red}.\:{you}\:{don}'{t}\:{care}\:{which}\:{red}\: \\ $$$${ball}\:{is}\:{taken},\:{because}\:{they}\:{are}\:{the} \\ $$$${same}\:{to}\:{you}.\:{but}\:{the}\:{balls}\:{themself} \\ $$$${care}.\:{there}\:{is}\:{difference}\:{if}\:{the}\:{ball} \\ $$$${A}\:{is}\:{taken}\:{or}\:{the}\:{ball}\:{B}\:{is}\:{taken}. \\ $$

Commented by mr W last updated on 13/Oct/22

for the same reason in the world of  probability   • • • , • • •  are two different things.

$${for}\:{the}\:{same}\:{reason}\:{in}\:{the}\:{world}\:{of} \\ $$$${probability} \\ $$$$\:\bullet\:\bullet\:\bullet\:,\:\bullet\:\bullet\:\bullet \\ $$$${are}\:{two}\:{different}\:{things}. \\ $$

Commented by Acem last updated on 13/Oct/22

 I back to the qusetion, it asks about    different results    hmmm.... i need a bit time to see if there′s   a difference between { ((how many result)),((how many different res.)) :}

$$\:{I}\:{back}\:{to}\:{the}\:{qusetion},\:{it}\:{asks}\:{about} \\ $$$$\:\:\boldsymbol{{different}}\:{results} \\ $$$$ \\ $$$${hmmm}....\:{i}\:{need}\:{a}\:{bit}\:{time}\:{to}\:{see}\:{if}\:{there}'{s} \\ $$$$\:{a}\:{difference}\:{between\begin{cases}{{how}\:{many}\:{result}}\\{{how}\:{many}\:{different}\:{res}.}\end{cases}} \\ $$$$ \\ $$

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