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

The number of ways of distributing six identical mathematics books and six identical physics books among three students such that each student gets atleast one mathematics book and atleast one physics book is ((5.5!)/k), then k is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{distributing}\:\mathrm{six} \\ $$$$\mathrm{identical}\:\mathrm{mathematics}\:\mathrm{books}\:\mathrm{and}\:\mathrm{six} \\ $$$$\mathrm{identical}\:\mathrm{physics}\:\mathrm{books}\:\mathrm{among}\:\mathrm{three} \\ $$$$\mathrm{students}\:\mathrm{such}\:\mathrm{that}\:\mathrm{each}\:\mathrm{student}\:\mathrm{gets} \\ $$$$\mathrm{atleast}\:\mathrm{one}\:\mathrm{mathematics}\:\mathrm{book}\:\mathrm{and} \\ $$$$\mathrm{atleast}\:\mathrm{one}\:\mathrm{physics}\:\mathrm{book}\:\mathrm{is}\:\frac{\mathrm{5}.\mathrm{5}!}{{k}},\:\mathrm{then}\:{k} \\ $$$$\mathrm{is} \\ $$

Commented by Tinkutara last updated on 07/Oct/17

Thank you very much Sir!

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

Commented by mrW1 last updated on 07/Oct/17

to distribute 6 mathe books among 3 students there are C_(3−1) ^(6−1) =C_2 ^5 ways to distribute 6 physics books among 3 students there are C_(3−1) ^(6−1) =C_2 ^5 ways ⇒total ways: C_2 ^5 ×C_2 ^5 =((5!)/(2!3!))×((5!)/(2!3!))=((5×5!×4×3!)/(2!2!3!3!)) =((5×5!×4)/(2!2!3!)) =((5×5!)/6) ⇒k=6

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

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