Find the number of ways in which a selection of $5$ books can be done out of $3$ physics, $3$ chemistry and $3$ mathematics books by taking at least one book of each subject, books of same subject being different.

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Solution

$\begin{matrix} T o t a l n u m b e r o f w a y s = b o o k s c a n b e t a k e n i n g r o u p o f (1, 3, 1) + I n g r o u p o f (1, 2, 2) f r o m m a t h s, p h y s i c s a n d c h e m i s t r y . = \frac{3!}{2!} (^{3} C_{1} \times^{3} C_{3} \times^{3} C_{1}) + \frac{3!}{2!} (^{3} C_{1} \times^{3} C_{3} \times^{3} C_{1}) = \frac{3 \times 1 \times 3 \times 3 \times 2}{2} + (3 \times 3 \times 3) \frac{6}{2} = 27 + 81 = 108 \end{matrix}$
Which, is the required answer.

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Find the number of ways in which a selection of 5 books can be done out of 3 physics, 3 chemistry and 3 mathematics books by taking at least one book of each subject, books of same subject being different.

Totalnumberofways=bookscanbetakeningroupof(1,3,1)+Ingroupof(1,2,2)frommaths,physicsandchemistry.=3!2!(3C1×3C3×3C1)+3!2!(3C1×3C3×3C1)=3×1×3×3×22+(3×3×3)62=27+81=108Which, is the required answer.

Find the number of ways in which a selection of $5$ books can be done out of $3$ physics, $3$ chemistry and $3$ mathematics books by taking at least one book of each subject, books of same subject being different.