A library has a copies of one book,b copies o | vedclass

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(B) The total number of books is $N = a + 2b + 3c + d$.The number of ways to arrange $N$ items where there are groups of identical items is given by t

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The total number of books is $N = a + 2b + 3c + d$.The number of ways to arrange $N$ items where there are groups of identical items is given by the multinomial coefficient formula: $\frac{N!}{n_1! n_2! ... n_k!}$.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 (each is distinct,so $1!$ for each,which is $1$).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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