Question

A library has 6 copies of one book, 4 copies of each of three books, and single copies of 8 books. Then the number of arrangements of all the
A. $\dfrac{{(26)!}}{{{{(4!)}^3}6!}}$
B. $\dfrac{{(26)!}}{{6!{{(4!)}^3}{{(6!)}^3}}}$
C. $\dfrac{{(26)!}}{{6!{{(4!)}^2}{{(6!)}^3}}}$
D. $\dfrac{{(26)!}}{{6!.4!.6!}}$

Answer 1

A library has a number of copies of one book, b copies each of three books, and a single copy of c books. The total number of ways in which these books can be arranged is equal to $\dfrac{{(a + 3b + c)!}}{{a!{{(b!)}^3}(c!)}}$

Complete step-by-step answer:
The library has 6 copies of one book, 4 copies of each of three books, and single copies of 8 books.
∴ Total no. of books $= 6 \times 1 + 4 \times 3 + 1 \times 8 = 6 + 12 + 8 = 26$
Since the library has 6 copies of one book, 4 copies of each of three books, and single copies of 8 books.
∴ Number of ways of arranging these books is $\dfrac{{(a + 3b + c)!}}{{a!{{(b!)}^3}(c!)}}$
Given, the value of a is 6, b is 4 and c is 1.
Putting these value in $\dfrac{{(a + 3b + c)!}}{{a!{{(b!)}^3}(c!)}}$, we get

$\Rightarrow \dfrac{{(6 + 3 \times 4 + 1)!}}{{6!{{(4!)}^3}(1!)}} \\\ \Rightarrow \dfrac{{(6 + 12 + 8)!}}{{6!{{(4!)}^3}(1!)}} \\\ \Rightarrow \dfrac{{(26)!}}{{{{(4!)}^3}6!}} \\\$
​

Hence the correct answer is $\dfrac{{(26)!}}{{{{(4!)}^3}6!}}$

So, the correct answer is “Option A”.

Note: We have to keep in mind that the number of ways of arranging n unlike objects in a line in n! For example, the different ways the P, Q, R, S can be arranged will be 4!
The number of ways of arranging n objects, of which p of one type are alike, q of a second type is alike, r of a third type are alike is $\dfrac{{n!}}{{p!q!r!}}$