Question

A library has 'a' copies of one book, 'b' copies each of two books, 'c' copies each of three books, and a single copy of 'd' books. The total number of ways in which these books can be arranged in a shelf is equal to

& A.\dfrac{\left( a+2b+3c+d \right)!}{a!{{\left( b! \right)}^{2}}{{\left( c! \right)}^{3}}} \\\ & B.\dfrac{\left( a+2b+3c+d \right)!}{a!\left( 2b! \right){{\left( c! \right)}^{3}}} \\\ & C.\dfrac{\left( a+b+3c+d \right)!}{{{\left( c! \right)}^{3}}} \\\ & D.\dfrac{\left( a+2b+3c+d \right)!}{a!\left( 2b \right)!\left( 3c \right)!} \\\ \end{aligned}$$

Answer 1

Find total number of books and find the number of ways to arrange them. For each book the number of copies are mentioned. Hence, divide the same from the total arrangements to get the number of ways to arrange books on the shelf.

Complete step by step answer:
It is said that a library has 'a' copies of one book ,'b' copies each of two books, 'c' copies each of three books, and a single copy of 'd' books.
Now we need to arrange these books on a shelf.
Hence, we can say it in simple terms as,
a copies - 1 book
b copies - 2 books
c copies - 3 books
1 copy - d books
Now we have to arrange all these books on a shelf. So let us find the total books.
$\text{Total Books}=\,(a\,\text{copies}\,\times \text{1}\,\text{book)}\,\text{+}(b\,\text{copies}\times \text{2}\,\text{books)+}(c\,\text{copies}\times \text{3}\,\text{book)+}(1\,\text{copy}\times \text{d}\,\text{books)}$

& \Rightarrow \text{a+2}\times \text{b}\,\text{+}\,\text{3}\times \text{c}\,\text{+}\,\text{d}\times 1 \\\ & \Rightarrow \,\text{a+2b}\,\text{+}\,\text{3c}\,\text{+}\,\text{d} \\\ \end{aligned}$$ Total number of books $$\Rightarrow \,\,\text{a+2b}\,\text{+}\,\text{3c}\,\text{+}\,\text{d}$$ Thus the number of ways of arranging $$\text{(a+2b}\,\text{+}\,\text{3c}\,\text{+}\,\text{d)}$$ books in the shelf $$\Rightarrow \text{(a+2b}\,\text{+}\,\text{3c}\,\text{+}\,\text{d})\,!\,----\left( \text{1} \right)$$ Now, on the book shelf, we said a copy of 1 book. Let us name it as book 1. Hence, there are copies of book 1.Then there are b copies of 2books. Let us name them as, book 2 and book 3.Both book 2 and book 3 has 'b' copies of books. Similarly, c copies of 3 books, then book 4, book 5 and book 6 each have 'c' copies. In simple words we can represent it as  The arrangement of the book doesn’t matter. The ’d’ books are distinct as there is only one single copy. Thus we have to divide the number of ways of arrangement by the above Total number of way $$\Rightarrow \dfrac{\left( a+2b+3c+d \right)!}{a!\,b!\,b!\,c!c\,!\,c!}\,\,\,=\,\,\dfrac{\left( a+2b+3c+d \right)!}{a!\,{{\left( b! \right)}^{2}}{{\left( c! \right)}^{3}}}$$ Thus the number of ways in which these books can be arranged in the shelf $$\,\Rightarrow \,\dfrac{\left( a+2b+3c+d \right)!}{a!\,{{\left( b! \right)}^{2}}{{\left( c! \right)}^{3}}}$$ **So, the correct answer is “Option A”.** **Note:** You only know that these are copies of books 1, 2, 3, 4, 5, 6 This arrangement of the books in order doesn’t matter. You can put the particular book anywhere on the shelf. You might take the total number of books as (a+b+c+d) which is wrong.