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(A) The total number of books is $N = a + 2b + 3c + d$. We have $a$ identical copies of one book,$b$ identical copies of each of two books,$c$ identic

Vedclass · Accepted Answer

$\frac{(a + 2b + 3c + d)!}{a! (b!)^2 (c!)^3}$

Vedclass · Answer

(A) The total number of books is $N = a + 2b + 3c + d$. We have $a$ identical copies of one book,$b$ identical copies of each of two books,$c$ identical copies of each of three books,and $d$ distinct single books. The number of arrangements of $N$ items where there are $n_1$ identical items of type $1$,$n_2$ identical items of type $2$,etc.,is given by $\frac{N!}{n_1! n_2! \dots}$. Here,we have: - $a$ identical copies of one book ($a!$ ways) - $b$ identical copies of each of two books ($(b!)^2$ ways) - $c$ identical copies of each of three books ($(c!)^3$ ways) - $d$ single copies of $d$ books ($1!$ ways for each,so $1^d = 1$) Thus,the total number of arrangements is $\frac{(a + 2b + 3c + d)!}{a! (b!)^2 (c!)^3}$.

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

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