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

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

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

$$ifpossiblepleasesolvegenerallyfor$$$$n_{m}mathematicsbooks$$$$n_{p}physicsbooks$$$$n_{c}chemistrybooks$$$$with{n}_m⩾n_p⩾n_c$$

Answered by bemath last updated on 04/Jul/20

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

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

$$pleaseexplainyouranswersir!$$

Commented by bemath last updated on 04/Jul/20

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

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

$$ihavenoansweryet.$$

Commented by bemath last updated on 04/Jul/20

$$\mathrm{hihihi}\dots$$

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

$$infollowing\mathit{M}isplaceholderfor$$$$amathematicsbook,\mathit{X},\mathit{Y}areplace$$$$holdersforphysicsorchemistry$$$$books.$$$$$$$$\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}$$$$11\times 2!\times \frac{10!}{6!4!}=4620$$$$\mathit{X}\mathit{Y}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}$$$$\mathit{M}\mathit{X}\mathit{Y}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}$$$$10\times (\frac{9!}{5!4!}+\frac{9!}{6!3!})\times 2=4200$$$$\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}$$$$\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}$$$$C_2^{10}\times 2\times 2\times \frac{8!}{5!3!}\times 2=20160$$$$\mathit{M}\mathit{X}\mathit{Y}\mathit{X}\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}$$$$9\times 8\times (2\times \frac{7!}{4!3!}+2\times \frac{7!}{5!2!})=8064$$$$\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{Y}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}\mathit{X}\mathit{M}$$$$C_3^9\times 2\times 2\times 2\times \frac{6!}{4!2!}=10080$$$$$$$$totally:$$$$4620+4200+20160+8064+10080$$$$=47124$$

Commented by bemath last updated on 04/Jul/20

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

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

$$theyaredifferent!we{can}^{\prime }tapply$$$$ageneralformulaforallcases.e.g.$$$$ifwehad10mathematicsbooks,$$$$4physicsbooksand4chemistrybooks,$$$$thereisevennosolution.$$

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