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Question Number 21931 by Tinkutara last updated on 07/Oct/17

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{distributing}\mathrm{six}$$$$\mathrm{identical}\mathrm{mathematics}\mathrm{books}\mathrm{and}\mathrm{six}$$$$\mathrm{identical}\mathrm{physics}\mathrm{books}\mathrm{among}\mathrm{three}$$$$\mathrm{students}\mathrm{such}\mathrm{that}\mathrm{each}\mathrm{student}\mathrm{gets}$$$$\mathrm{atleast}\mathrm{one}\mathrm{mathematics}\mathrm{book}\mathrm{and}$$$$\mathrm{atleast}\mathrm{one}\mathrm{physics}\mathrm{book}\mathrm{is}\frac{5.5!}{k},\mathrm{then}k$$$$\mathrm{is}$$

Commented by Tinkutara last updated on 07/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by mrW1 last updated on 07/Oct/17

$$\mathrm{to}\mathrm{distribute}6\mathrm{mathe}\mathrm{books}\mathrm{among}$$$$3\mathrm{students}\mathrm{there}\mathrm{are}{\mathrm{C}}_{3-1}^{6-1}={\mathrm{C}}_2^5\mathrm{ways}$$$$$$$$\mathrm{to}\mathrm{distribute}6\mathrm{physics}\mathrm{books}\mathrm{among}$$$$3\mathrm{students}\mathrm{there}\mathrm{are}{\mathrm{C}}_{3-1}^{6-1}={\mathrm{C}}_2^5\mathrm{ways}$$$$$$$$\Rightarrow \mathrm{total}\mathrm{ways}:$$$${\mathrm{C}}_2^5\times {\mathrm{C}}_2^5=\frac{5!}{2!3!}\times \frac{5!}{2!3!}=\frac{5\times 5!\times 4\times 3!}{2!2!3!3!}$$$$=\frac{5\times 5!\times 4}{2!2!3!}$$$$=\frac{5\times 5!}{6}$$$$\Rightarrow \mathrm{k}=6$$

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