Menu Close

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




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

Leave a Reply

Your email address will not be published. Required fields are marked *