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Question Number 101693 by bemath last updated on 04/Jul/20

There are 4 identical mathematics books, 2 identic physics books and 2 identical chemistry books . How many ways to compile the eight books on the condition of the same book are not mutually adjacent?

$$\mathrm{There}\:\mathrm{are}\:\mathrm{4}\:\mathrm{identical}\:\mathrm{mathematics} \\ $$$$\mathrm{books},\:\mathrm{2}\:\mathrm{identic}\:\mathrm{physics}\:\mathrm{books} \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{identical}\:\mathrm{chemistry}\:\mathrm{books} \\ $$$$.\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{compile}\: \\ $$$$\mathrm{the}\:\mathrm{eight}\:\mathrm{books}\:\mathrm{on}\:\mathrm{the}\:\mathrm{condition} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{book}\:\mathrm{are}\:\mathrm{not}\:\mathrm{mutually} \\ $$$$\mathrm{adjacent}? \\ $$

Commented by bobhans last updated on 04/Jul/20

MPMPMCMC, MPMCMPMC, MPMCMCMP, MCMPMPMC, MCMPMCMP, MCMCMPMP, PMPMCMCM, PMCMPMCM, PMCMCMPM, CMPMPMCM, CMPMCMPM, CMCMPMPM, MPMCMPCM, MPMCMCPM, MCMPMPCM, MCMPMCPM, MPMPCMCM, MPMCPMCM, MCMPCMPM, MCMCPMPM, MPCMPMCM, MCPMPMCM, MPCMCMPM, MCPMCMPM

Commented by bemath last updated on 04/Jul/20

what the formula?

$$\mathrm{what}\:\mathrm{the}\:\mathrm{formula}? \\ $$

Answered by bobhans last updated on 04/Jul/20

24 ways

$$\mathrm{24}\:\mathrm{ways} \\ $$

Commented by bobhans last updated on 04/Jul/20

M_M_M_M_ ⇔ P_4 ^4 = ((4!)/((4−4)!)) = 24 ★

$$\mathrm{M\_M\_M\_M\_}\:\Leftrightarrow\:\mathrm{P}_{\mathrm{4}} ^{\mathrm{4}} \:=\:\frac{\mathrm{4}!}{\left(\mathrm{4}−\mathrm{4}\right)!}\:=\:\mathrm{24}\:\bigstar \\ $$

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

can you give some explanation to the formula sir?

$${can}\:{you}\:{give}\:{some}\:{explanation}\:{to} \\ $$$${the}\:{formula}\:{sir}? \\ $$

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

case 1: XMXMXMXM ⇒((4!)/(2!2!))=6 case 2: MXMXMXMX ⇒((4!)/(2!2!))=6 case 3: MXYMXMXM, MXMXYMXM, MXMXMXYM ⇒3×2×2=12 totally: 6+6+12=24

$${case}\:\mathrm{1}:\:{XMXMXMXM} \\ $$$$\Rightarrow\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{6} \\ $$$${case}\:\mathrm{2}:\:{MXMXMXMX} \\ $$$$\Rightarrow\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{6} \\ $$$${case}\:\mathrm{3}:\:{MXYMXMXM},\:{MXMXYMXM},\:{MXMXMXYM} \\ $$$$\Rightarrow\mathrm{3}×\mathrm{2}×\mathrm{2}=\mathrm{12} \\ $$$$ \\ $$$${totally}:\:\mathrm{6}+\mathrm{6}+\mathrm{12}=\mathrm{24} \\ $$

Answered by john santu last updated on 05/Jul/20

If the math books are in position 1/3/5/7 or 2/4/6/8 there are 6 ways to arrange the physics and the chemistry → 2×6 = 12 but if the math books are in position 1/3/5/8 or 1/3/6/8 or 1/4/6/8 then there are only 4 way → 3×4 =12 so totally = 24 ways (JS ⊛)

$${If}\:{the}\:{math}\:{books}\:{are}\:{in}\:{position} \\ $$$$\mathrm{1}/\mathrm{3}/\mathrm{5}/\mathrm{7}\:{or}\:\mathrm{2}/\mathrm{4}/\mathrm{6}/\mathrm{8}\:{there}\:{are} \\ $$$$\mathrm{6}\:{ways}\:{to}\:{arrange}\:{the}\:{physics} \\ $$$${and}\:{the}\:{chemistry}\:\rightarrow\:\mathrm{2}×\mathrm{6}\:=\:\mathrm{12} \\ $$$${but}\:{if}\:{the}\:{math}\:{books}\:{are}\:{in} \\ $$$${position}\:\mathrm{1}/\mathrm{3}/\mathrm{5}/\mathrm{8}\:{or}\:\mathrm{1}/\mathrm{3}/\mathrm{6}/\mathrm{8}\: \\ $$$${or}\:\mathrm{1}/\mathrm{4}/\mathrm{6}/\mathrm{8}\:{then}\:{there}\:{are}\:{only} \\ $$$$\mathrm{4}\:{way}\:\rightarrow\:\mathrm{3}×\mathrm{4}\:=\mathrm{12} \\ $$$${so}\:{totally}\:=\:\mathrm{24}\:{ways}\:\:\left({JS}\:\circledast\right)\: \\ $$

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