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Question Number 101732 by mr W last updated on 04/Jul/20

There are 10 identical mathematics books, 7 identical physics books and 5 identical chemistry books. Find the number of ways to compile the books under the condition that same books are not mutually adjacent.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{10}\:\mathrm{identical}\:\mathrm{mathematics} \\ $$$$\mathrm{books},\:\mathrm{7}\:\mathrm{identical}\:\mathrm{physics}\:\mathrm{books} \\ $$$$\mathrm{and}\:\mathrm{5}\:\mathrm{identical}\:\mathrm{chemistry}\:\mathrm{books}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{compile}\: \\ $$$$\mathrm{the}\:\mathrm{books}\:\mathrm{under}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{that} \\ $$$$\mathrm{same}\:\mathrm{books}\:\mathrm{are}\:\mathrm{not}\:\mathrm{mutually}\:\mathrm{adjacent}. \\ $$

Commented by mr W last updated on 04/Jul/20

if possible please solve generally for n_m mathematics books n_p physics books n_c chemistry books with n_m ≥n_p ≥n_c

$${if}\:{possible}\:{please}\:{solve}\:{generally}\:{for} \\ $$$${n}_{{m}} \:{mathematics}\:{books} \\ $$$${n}_{{p}} \:{physics}\:{books} \\ $$$${n}_{{c}} \:{chemistry}\:{books} \\ $$$${with}\:{n}_{{m}} \geqslant{n}_{{p}} \geqslant{n}_{{c}} \\ $$

Answered by bemath last updated on 04/Jul/20

case(1) 2×((12!)/(7!.5!)) case(2)9×7×5 totally = ((2×12!)/(7!.5!)) + 9×7×5

$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{2}×\frac{\mathrm{12}!}{\mathrm{7}!.\mathrm{5}!} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\mathrm{9}×\mathrm{7}×\mathrm{5} \\ $$$$\mathrm{totally}\:=\:\frac{\mathrm{2}×\mathrm{12}!}{\mathrm{7}!.\mathrm{5}!}\:+\:\mathrm{9}×\mathrm{7}×\mathrm{5} \\ $$$$ \\ $$

Commented by mr W last updated on 04/Jul/20

please explain your answer sir!

$${please}\:{explain}\:{your}\:{answer}\:{sir}! \\ $$

Commented by bemath last updated on 04/Jul/20

wrong or correct sir?

$$\mathrm{wrong}\:\mathrm{or}\:\mathrm{correct}\:\mathrm{sir}?\: \\ $$

Commented by mr W last updated on 04/Jul/20

i have no answer yet.

$${i}\:{have}\:{no}\:{answer}\:{yet}. \\ $$

Commented by bemath last updated on 04/Jul/20

hihihi...

$$\mathrm{hihihi}... \\ $$

Answered by mr W last updated on 04/Jul/20

in following M is place holder for a mathematics book, X, Y are place holders for physics or chemistry books. XYMXMXMXMXMXMXMXMXMXMX 11×2!×((10!)/(6!4!))=4620 XYXMXMXMXMXMXMXMXMXMXM MXYXMXMXMXMXMXMXMXMXMX 10×(((9!)/(5!4!))+((9!)/(6!3!)))×2=4200 XYMXYMXMXMXMXMXMXMXMXM MXYMXYMXMXMXMXMXMXMXMX C_2 ^(10) ×2×2×((8!)/(5!3!))×2=20160 MXYXMXYMXMXMXMXMXMXMXM 9×8×(2×((7!)/(4!3!))+2×((7!)/(5!2!)))=8064 MXYMXYMXYMXMXMXMXMXMXM C_3 ^9 ×2×2×2×((6!)/(4!2!))=10080 totally: 4620+4200+20160+8064+10080 =47124

$${in}\:{following}\:\boldsymbol{{M}}\:{is}\:{place}\:{holder}\:{for} \\ $$$${a}\:{mathematics}\:{book},\:\boldsymbol{{X}},\:\boldsymbol{{Y}}\:\:{are}\:{place} \\ $$$${holders}\:{for}\:{physics}\:{or}\:{chemistry} \\ $$$${books}. \\ $$$$ \\ $$$$\boldsymbol{{XYMXMXMXMXMXMXMXMXMXMX}} \\ $$$$\mathrm{11}×\mathrm{2}!×\frac{\mathrm{10}!}{\mathrm{6}!\mathrm{4}!}=\mathrm{4620} \\ $$$$\boldsymbol{{XYXMXMXMXMXMXMXMXMXMXM}} \\ $$$$\boldsymbol{{MXYXMXMXMXMXMXMXMXMXMX}} \\ $$$$\mathrm{10}×\left(\frac{\mathrm{9}!}{\mathrm{5}!\mathrm{4}!}+\frac{\mathrm{9}!}{\mathrm{6}!\mathrm{3}!}\right)×\mathrm{2}=\mathrm{4200} \\ $$$$\boldsymbol{{XYMXYMXMXMXMXMXMXMXMXM}} \\ $$$$\boldsymbol{{MXYMXYMXMXMXMXMXMXMXMX}} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{10}} ×\mathrm{2}×\mathrm{2}×\frac{\mathrm{8}!}{\mathrm{5}!\mathrm{3}!}×\mathrm{2}=\mathrm{20160} \\ $$$$\boldsymbol{{MXYXMXYMXMXMXMXMXMXMXM}} \\ $$$$\mathrm{9}×\mathrm{8}×\left(\mathrm{2}×\frac{\mathrm{7}!}{\mathrm{4}!\mathrm{3}!}+\mathrm{2}×\frac{\mathrm{7}!}{\mathrm{5}!\mathrm{2}!}\right)=\mathrm{8064} \\ $$$$\boldsymbol{{MXYMXYMXYMXMXMXMXMXMXM}} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{9}} ×\mathrm{2}×\mathrm{2}×\mathrm{2}×\frac{\mathrm{6}!}{\mathrm{4}!\mathrm{2}!}=\mathrm{10080} \\ $$$$ \\ $$$${totally}: \\ $$$$\mathrm{4620}+\mathrm{4200}+\mathrm{20160}+\mathrm{8064}+\mathrm{10080} \\ $$$$=\mathrm{47124} \\ $$

Commented by bemath last updated on 04/Jul/20

why sir different to eq 101693

$$\mathrm{why}\:\mathrm{sir}\:\mathrm{different}\:\mathrm{to}\:\mathrm{eq}\:\mathrm{101693} \\ $$

Commented by mr W last updated on 04/Jul/20

they are different! we can′t apply a general formula for all cases. e.g. if we had 10 mathematics books, 4 physics books and 4 chemistry books, there is even no solution.

$${they}\:{are}\:{different}!\:{we}\:{can}'{t}\:{apply} \\ $$$${a}\:{general}\:{formula}\:{for}\:{all}\:{cases}.\:{e}.{g}. \\ $$$${if}\:{we}\:{had}\:\mathrm{10}\:{mathematics}\:{books}, \\ $$$$\mathrm{4}\:{physics}\:{books}\:{and}\:\mathrm{4}\:{chemistry}\:{books}, \\ $$$${there}\:{is}\:{even}\:{no}\:{solution}. \\ $$

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